Samacheer Kalvi Class 11 Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.1

Question 1. Find the derivatives of the following functions using the first principle.
(i) f(x) = 6
(ii) f(x) = -4x + 7
(iii) f(x) = -x² + 2
Answer:
To find the derivative using the first principle, we use the formula: \( f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \).

(i) For \( f(x) = 6 \):
First, find \( f(x + \Delta x) \). Since \( f(x) \) is a constant function, \( f(x + \Delta x) = 6 \).
Next, calculate the difference: \( f(x + \Delta x) - f(x) = 6 - 6 = 0 \).
Now, apply the limit: \( f'(x) = \lim_{\Delta x \to 0} \frac{0}{\Delta x} = 0 \).
So, the derivative of \( f(x) = 6 \) is \( 0 \). The derivative of any constant is always zero.

(ii) For \( f(x) = -4x + 7 \):
First, find \( f(x + \Delta x) = -4(x + \Delta x) + 7 = -4x - 4\Delta x + 7 \).
Next, calculate the difference: \( f(x + \Delta x) - f(x) = (-4x - 4\Delta x + 7) - (-4x + 7) \).
\( = -4x - 4\Delta x + 7 + 4x - 7 = -4\Delta x \).
Now, apply the limit: \( f'(x) = \lim_{\Delta x \to 0} \frac{-4\Delta x}{\Delta x} = \lim_{\Delta x \to 0} (-4) = -4 \).
So, the derivative of \( f(x) = -4x + 7 \) is \( -4 \). This is consistent with the power rule for derivatives.

(iii) For \( f(x) = -x^2 + 2 \):
First, find \( f(x + \Delta x) = -(x + \Delta x)^2 + 2 = -(x^2 + 2x\Delta x + (\Delta x)^2) + 2 \).
Next, calculate the difference: \( f(x + \Delta x) - f(x) = [-(x^2 + 2x\Delta x + (\Delta x)^2) + 2] - [-x^2 + 2] \).
\( = -x^2 - 2x\Delta x - (\Delta x)^2 + 2 + x^2 - 2 = -2x\Delta x - (\Delta x)^2 \).
Now, apply the limit: \( f'(x) = \lim_{\Delta x \to 0} \frac{-2x\Delta x - (\Delta x)^2}{\Delta x} \).
\( = \lim_{\Delta x \to 0} (-2x - \Delta x) \).
Substitute \( \Delta x = 0 \): \( f'(x) = -2x - 0 = -2x \).
So, the derivative of \( f(x) = -x^2 + 2 \) is \( -2x \). The term \( (\Delta x)^2 \) becomes zero in the limit as \( \Delta x \) approaches zero.
In simple words: To find the derivative using the first principle, we look at how much the function's value changes as 'x' changes by a tiny bit. Then, we divide this change by the tiny change in 'x' and see what happens as that change gets smaller and smaller, almost zero. For (i), a constant function doesn't change, so its derivative is zero. For (ii) and (iii), we expand the function with \( \Delta x \), simplify the difference, and then take the limit.

🎯 Exam Tip: Remember the definition of the derivative from the first principle. Simplify the expression \( \frac{f(x + \Delta x) - f(x)}{\Delta x} \) as much as possible before taking the limit to avoid indeterminate forms.

Question 2. Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?
(i) f(x) = |x - 1|
(ii) f(x) = \( \sqrt{1 - x^2} \)
Answer:
(i) For \( f(x) = |x - 1| \) at \( x = 1 \):
First, we define \( f(x) \) as a piecewise function:
If \( x - 1 > 0 \), i.e., \( x > 1 \), then \( f(x) = x - 1 \).
If \( x - 1 < 0 \), i.e., \( x < 1 \), then \( f(x) = -(x - 1) \).
Also, \( f(1) = |1 - 1| = 0 \).

Left-hand derivative at \( x = 1 \):
\( f'(1^-) = \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} \)
Since \( x \to 1^- \), we use \( f(x) = -(x - 1) \).
\( f'(1^-) = \lim_{x \to 1^-} \frac{-(x - 1) - 0}{x - 1} \)
\( = \lim_{x \to 1^-} (-1) = -1 \).

Right-hand derivative at \( x = 1 \):
\( f'(1^+) = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} \)
Since \( x \to 1^+ \), we use \( f(x) = x - 1 \).
\( f'(1^+) = \lim_{x \to 1^+} \frac{(x - 1) - 0}{x - 1} \)
\( = \lim_{x \to 1^+} (1) = 1 \).

Since the left-hand derivative (\( -1 \)) is not equal to the right-hand derivative (\( 1 \)), \( f(x) = |x - 1| \) is not differentiable at \( x = 1 \). Functions with sharp "V" shapes at a point are generally not differentiable there.

(ii) For \( f(x) = \sqrt{1 - x^2} \) at \( x = 1 \):
First, find \( f(1) = \sqrt{1 - 1^2} = \sqrt{0} = 0 \).

Left-hand derivative at \( x = 1 \):
Let \( x = 1 - h \), where \( h \to 0^+ \).
\( f'(1^-) = \lim_{h \to 0^+} \frac{f(1 - h) - f(1)}{(1 - h) - 1} \)
\( = \lim_{h \to 0^+} \frac{\sqrt{1 - (1 - h)^2} - 0}{-h} \)
\( = \lim_{h \to 0^+} \frac{\sqrt{1 - (1 - 2h + h^2)}}{-h} \)
\( = \lim_{h \to 0^+} \frac{\sqrt{2h - h^2}}{-h} = \lim_{h \to 0^+} \frac{\sqrt{h(2 - h)}}{-h} \)
\( = \lim_{h \to 0^+} \frac{\sqrt{h}\sqrt{2 - h}}{-\sqrt{h}\sqrt{h}} = \lim_{h \to 0^+} \frac{\sqrt{2 - h}}{-\sqrt{h}} \).
As \( h \to 0^+ \), the numerator approaches \( \sqrt{2} \), and the denominator approaches \( 0 \) from the negative side. This results in \( -\infty \).
So, \( f'(1^-) = -\infty \).

Right-hand derivative at \( x = 1 \):
Let \( x = 1 + h \), where \( h \to 0^+ \).
\( f'(1^+) = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{(1 + h) - 1} \)
\( = \lim_{h \to 0^+} \frac{\sqrt{1 - (1 + h)^2} - 0}{h} \)
\( = \lim_{h \to 0^+} \frac{\sqrt{1 - (1 + 2h + h^2)}}{h} \)
\( = \lim_{h \to 0^+} \frac{\sqrt{-2h - h^2}}{h} \).
For \( h > 0 \), the term inside the square root, \( -2h - h^2 = -h(2 + h) \), is negative. The square root of a negative number is not a real number. This means the function \( f(x) \) is not defined for values greater than \( 1 \), so the right-hand derivative does not exist.

Since the left-hand derivative is infinite and the right-hand derivative does not exist in real numbers, the function \( f(x) = \sqrt{1 - x^2} \) is not differentiable at \( x = 1 \). This function describes the upper half of a circle, which has a vertical tangent at \( x = 1 \), hence an infinite slope.
In simple words: To check if a function is differentiable at a point, we calculate the derivative from the left side and the right side of that point. If these two values are different or do not exist, the function is not differentiable. For \( |x-1| \), the graph has a sharp corner at \( x=1 \), so it's not smooth enough to have a single derivative. For \( \sqrt{1-x^2} \), the function only exists for \( x \) values between -1 and 1, forming a semi-circle. At \( x=1 \), the tangent line is vertical, meaning the slope is infinite, so the derivative doesn't exist.

🎯 Exam Tip: Functions involving absolute values or square roots often have points where they are not differentiable due to sharp corners or vertical tangents. Always check the domain of the function carefully, especially for square root expressions.

Question 3. Determine whether the following function is differentiable at the indicated values.
(i) f(x) = x |x| at x = 0
(ii) f(x) = |x² - 1| at x = 1
(iii) f(x) = |x| + |x - 1| at x = 0, 1
(iv) f(x) = sin |x| at x = 0
Answer:
(i) For \( f(x) = x|x| \) at \( x = 0 \):
First, express \( f(x) \) as a piecewise function:
If \( x < 0 \), then \( |x| = -x \), so \( f(x) = x(-x) = -x^2 \).
If \( x \ge 0 \), then \( |x| = x \), so \( f(x) = x(x) = x^2 \).
At \( x = 0 \), \( f(0) = 0|0| = 0 \).

Left-hand derivative at \( x = 0 \):
\( f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} \)
Since \( x \to 0^- \), use \( f(x) = -x^2 \).
\( f'(0^-) = \lim_{x \to 0^-} \frac{-x^2 - 0}{x} = \lim_{x \to 0^-} (-x) = 0 \).

Right-hand derivative at \( x = 0 \):
\( f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} \)
Since \( x \to 0^+ \), use \( f(x) = x^2 \).
\( f'(0^+) = \lim_{x \to 0^+} \frac{x^2 - 0}{x} = \lim_{x \to 0^+} (x) = 0 \).

Since \( f'(0^-) = f'(0^+) = 0 \), the function \( f(x) = x|x| \) is differentiable at \( x = 0 \). This function is smooth at the origin.

(ii) For \( f(x) = |x^2 - 1| \) at \( x = 1 \):
First, express \( f(x) \) as a piecewise function based on the sign of \( x^2 - 1 \):
If \( x^2 - 1 > 0 \), i.e., \( x < -1 \) or \( x > 1 \), then \( f(x) = x^2 - 1 \).
If \( x^2 - 1 < 0 \), i.e., \( -1 < x < 1 \), then \( f(x) = -(x^2 - 1) = 1 - x^2 \).
At \( x = 1 \), \( f(1) = |1^2 - 1| = 0 \).

Left-hand derivative at \( x = 1 \):
\( f'(1^-) = \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} \)
Since \( x \to 1^- \), we consider \( -1 < x < 1 \), so \( f(x) = 1 - x^2 \).
\( f'(1^-) = \lim_{x \to 1^-} \frac{(1 - x^2) - 0}{x - 1} = \lim_{x \to 1^-} \frac{-(x^2 - 1)}{x - 1} \)
\( = \lim_{x \to 1^-} \frac{-(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1^-} -(x + 1) = -(1 + 1) = -2 \).

Right-hand derivative at \( x = 1 \):
\( f'(1^+) = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} \)
Since \( x \to 1^+ \), we consider \( x > 1 \), so \( f(x) = x^2 - 1 \).
\( f'(1^+) = \lim_{x \to 1^+} \frac{(x^2 - 1) - 0}{x - 1} = \lim_{x \to 1^+} \frac{(x - 1)(x + 1)}{x - 1} \)
\( = \lim_{x \to 1^+} (x + 1) = (1 + 1) = 2 \).

Since \( f'(1^-) \neq f'(1^+) \), the function \( f(x) = |x^2 - 1| \) is not differentiable at \( x = 1 \). This function also has a sharp corner at \( x=1 \).

(iii) For \( f(x) = |x| + |x - 1| \) at \( x = 0 \) and \( x = 1 \):
First, express \( f(x) \) as a piecewise function:
Case 1: \( x < 0 \). Then \( |x| = -x \) and \( |x - 1| = -(x - 1) = 1 - x \).
So, \( f(x) = -x + (1 - x) = 1 - 2x \).
Case 2: \( 0 \le x < 1 \). Then \( |x| = x \) and \( |x - 1| = -(x - 1) = 1 - x \).
So, \( f(x) = x + (1 - x) = 1 \).
Case 3: \( x \ge 1 \). Then \( |x| = x \) and \( |x - 1| = x - 1 \).
So, \( f(x) = x + (x - 1) = 2x - 1 \).

**Differentiability at \( x = 0 \):**
\( f(0) = |0| + |0 - 1| = 0 + 1 = 1 \).
Left-hand derivative at \( x = 0 \):
\( f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} \)
Using Case 1, \( f(x) = 1 - 2x \).
\( f'(0^-) = \lim_{x \to 0^-} \frac{(1 - 2x) - 1}{x} = \lim_{x \to 0^-} \frac{-2x}{x} = \lim_{x \to 0^-} (-2) = -2 \).
Right-hand derivative at \( x = 0 \):
\( f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} \)
Using Case 2, \( f(x) = 1 \).
\( f'(0^+) = \lim_{x \to 0^+} \frac{1 - 1}{x} = \lim_{x \to 0^+} \frac{0}{x} = 0 \).
Since \( f'(0^-) \neq f'(0^+) \), \( f(x) \) is not differentiable at \( x = 0 \).

**Differentiability at \( x = 1 \):**
\( f(1) = |1| + |1 - 1| = 1 + 0 = 1 \).
Left-hand derivative at \( x = 1 \):
\( f'(1^-) = \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} \)
Using Case 2, \( f(x) = 1 \).
\( f'(1^-) = \lim_{x \to 1^-} \frac{1 - 1}{x - 1} = \lim_{x \to 1^-} \frac{0}{x - 1} = 0 \).
Right-hand derivative at \( x = 1 \):
\( f'(1^+) = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} \)
Using Case 3, \( f(x) = 2x - 1 \).
\( f'(1^+) = \lim_{x \to 1^+} \frac{(2x - 1) - 1}{x - 1} = \lim_{x \to 1^+} \frac{2x - 2}{x - 1} \)
\( = \lim_{x \to 1^+} \frac{2(x - 1)}{x - 1} = \lim_{x \to 1^+} (2) = 2 \).
Since \( f'(1^-) \neq f'(1^+) \), \( f(x) \) is not differentiable at \( x = 1 \).

(iv) For \( f(x) = \sin|x| \) at \( x = 0 \):
First, express \( f(x) \) as a piecewise function:
If \( x < 0 \), then \( |x| = -x \), so \( f(x) = \sin(-x) = -\sin x \).
If \( x \ge 0 \), then \( |x| = x \), so \( f(x) = \sin x \).
At \( x = 0 \), \( f(0) = \sin|0| = \sin 0 = 0 \).

Left-hand derivative at \( x = 0 \):
\( f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} \)
Using \( f(x) = -\sin x \).
\( f'(0^-) = \lim_{x \to 0^-} \frac{-\sin x - 0}{x} = \lim_{x \to 0^-} \left(-\frac{\sin x}{x}\right) \).
Using the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we get \( f'(0^-) = -1 \).

Right-hand derivative at \( x = 0 \):
\( f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} \)
Using \( f(x) = \sin x \).
\( f'(0^+) = \lim_{x \to 0^+} \frac{\sin x - 0}{x} = \lim_{x \to 0^+} \frac{\sin x}{x} \).
Using the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we get \( f'(0^+) = 1 \).

Since \( f'(0^-) \neq f'(0^+) \), the function \( f(x) = \sin|x| \) is not differentiable at \( x = 0 \). The absolute value around \( x \) creates a sharp corner for \( \sin|x| \) at \( x=0 \).
In simple words: We check if a function can be smoothly differentiated at a point by comparing its left and right derivatives. If these derivatives are different, the function is not smooth enough at that point and cannot be differentiated there. For \( x|x| \), the derivatives match, so it's differentiable. For \( |x^2-1| \), \( |x|+|x-1| \), and \( \sin|x| \), the left and right derivatives are different, meaning they have sharp points or corners where the derivative does not exist.

🎯 Exam Tip: When dealing with absolute value functions, always rewrite them as piecewise functions. This helps in correctly identifying the function's form for left-hand and right-hand limits. Also, remember standard limits like \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

Question 4. Show that the following functions are not differentiable at the indicated value of x.
(i) f(x) = \( \begin{cases} -x + 2 & x \le 2 \\ 2x - 4 & x > 2 \end{cases} \) at x = 2
(ii) f(x) = \( \begin{cases} 3x & x < 0 \\ -4x & x \ge 0 \end{cases} \) at x = 0
Answer:
(i) For \( f(x) = \begin{cases} -x + 2 & x \le 2 \\ 2x - 4 & x > 2 \end{cases} \) at \( x = 2 \):
First, find the function value at \( x = 2 \). Since \( x \le 2 \), use the first rule:
\( f(2) = -2 + 2 = 0 \).

Left-hand derivative at \( x = 2 \):
\( f'(2^-) = \lim_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} \)
Since \( x \to 2^- \), we use \( f(x) = -x + 2 \).
\( f'(2^-) = \lim_{x \to 2^-} \frac{(-x + 2) - 0}{x - 2} = \lim_{x \to 2^-} \frac{-(x - 2)}{x - 2} \)
\( = \lim_{x \to 2^-} (-1) = -1 \).

Right-hand derivative at \( x = 2 \):
\( f'(2^+) = \lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} \)
Since \( x \to 2^+ \), we use \( f(x) = 2x - 4 \).
\( f'(2^+) = \lim_{x \to 2^+} \frac{(2x - 4) - 0}{x - 2} = \lim_{x \to 2^+} \frac{2(x - 2)}{x - 2} \)
\( = \lim_{x \to 2^+} (2) = 2 \).

Since \( f'(2^-) = -1 \) and \( f'(2^+) = 2 \), the left-hand derivative is not equal to the right-hand derivative. Therefore, \( f(x) \) is not differentiable at \( x = 2 \). The function changes its behavior abruptly at \( x=2 \).

(ii) For \( f(x) = \begin{cases} 3x & x < 0 \\ -4x & x \ge 0 \end{cases} \) at \( x = 0 \):
First, find the function value at \( x = 0 \). Since \( x \ge 0 \), use the second rule:
\( f(0) = -4 \times 0 = 0 \).

Left-hand derivative at \( x = 0 \):
\( f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} \)
Since \( x \to 0^- \), we use \( f(x) = 3x \).
\( f'(0^-) = \lim_{x \to 0^-} \frac{3x - 0}{x} = \lim_{x \to 0^-} (3) = 3 \).

Right-hand derivative at \( x = 0 \):
\( f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} \)
Since \( x \to 0^+ \), we use \( f(x) = -4x \).
\( f'(0^+) = \lim_{x \to 0^+} \frac{-4x - 0}{x} = \lim_{x \to 0^+} (-4) = -4 \).

Since \( f'(0^-) = 3 \) and \( f'(0^+) = -4 \), the left-hand derivative is not equal to the right-hand derivative. Therefore, \( f(x) \) is not differentiable at \( x = 0 \). This type of function often creates a sharp corner at the point where the definition changes.
In simple words: To show that a function is not differentiable at a certain point, we need to prove that its left-hand derivative and right-hand derivative at that point are not the same. For piecewise functions, this often happens at the points where the function's definition changes. If the slopes from both sides are different, the function has a sharp corner, not a smooth curve, so it's not differentiable.

🎯 Exam Tip: For piecewise functions, the points where the definition changes are critical for differentiability tests. Always calculate both the left and right derivatives at these points to see if they match.

Question 4. Show that the following functions are not differentiable at the indicated value of X
(i) \( f(x) = \begin{cases} -x + 2 & ; x \le 2 \\ 2x - 4 & ; x > 2 \end{cases} \) at \( x = 2 \)
Answer:
To check if \( f(x) \) is differentiable at \( x = 2 \), we need to find the left-hand derivative and the right-hand derivative at \( x = 2 \). If they are equal, the function is differentiable.
First, find the left-hand limit of \( f(x) \) at \( x = 2 \).
When \( x < 2 \), we use \( f(x) = -x + 2 \).
So, \( f(2) = -2 + 2 = 0 \).
The left-hand derivative, \( f'(2^-) \), is calculated as:
\( f'(2^-) = \lim_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} \)
\( = \lim_{x \to 2^-} \frac{(-x + 2) - 0}{x - 2} \)
\( = \lim_{x \to 2^-} \frac{-(x - 2)}{x - 2} \)
\( = \lim_{x \to 2^-} (-1) \)
\( = -1 \) -----(1)
Next, find the right-hand limit of \( f(x) \) at \( x = 2 \).
When \( x > 2 \), we use \( f(x) = 2x - 4 \).
So, \( f(2) = 2 \times 2 - 4 = 4 - 4 = 0 \).
The right-hand derivative, \( f'(2^+) \), is calculated as:
\( f'(2^+) = \lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} \)
\( = \lim_{x \to 2^+} \frac{(2x - 4) - 0}{x - 2} \)
\( = \lim_{x \to 2^+} \frac{2(x - 2)}{x - 2} \)
\( = \lim_{x \to 2^+} (2) \)
\( = 2 \) -----(2)
From equations (1) and (2), we see that \( f'(2^-) \neq f'(2^+) \) because \( -1 \neq 2 \). Since the left-hand and right-hand derivatives are not equal, the function \( f(x) \) is not differentiable at \( x = 2 \).
In simple words: A function is differentiable at a point if its slope is the same when you approach that point from the left and from the right. Here, the slopes are different at \( x = 2 \), so the function has a sharp corner and cannot be differentiated at that point.

🎯 Exam Tip: Always calculate both the left-hand and right-hand derivatives using the first principle definition to determine differentiability at a specific point, especially for piecewise functions.

(ii) \( f(x) = \begin{cases} 3x & ; x < 0 \\ -4x & ; x \ge 0 \end{cases} \) at \( x = 0 \)
Answer:
To check if \( f(x) \) is differentiable at \( x = 0 \), we must find the left-hand and right-hand derivatives at this point.
First, let's find the left-hand limit of \( f(x) \) at \( x = 0 \).
When \( x < 0 \), we use \( f(x) = 3x \).
So, \( f(0) = 3 \times 0 = 0 \).
The left-hand derivative, \( f'(0^-) \), is calculated as:
\( f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} \)
\( = \lim_{x \to 0^-} \frac{3x - 0}{x} \)
\( = \lim_{x \to 0^-} (3) \)
\( = 3 \) -----(1)
Next, let's find the right-hand limit of \( f(x) \) at \( x = 0 \).
When \( x \ge 0 \), we use \( f(x) = -4x \).
So, \( f(0) = -4 \times 0 = 0 \).
The right-hand derivative, \( f'(0^+) \), is calculated as:
\( f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} \)
\( = \lim_{x \to 0^+} \frac{-4x - 0}{x} \)
\( = \lim_{x \to 0^+} (-4) \)
\( = -4 \) -----(2)
From equations (1) and (2), we see that \( f'(0^-) \neq f'(0^+) \) because \( 3 \neq -4 \). Since the left-hand and right-hand derivatives are not equal, the function \( f(x) \) is not differentiable at \( x = 0 \). This function also has a sharp change in direction at \( x=0 \).
In simple words: The function's slope changes abruptly at \( x = 0 \). Approaching from the left, the slope is 3, but from the right, it's -4. Because these slopes are different, the function is not smooth at \( x = 0 \) and cannot be differentiated there.

🎯 Exam Tip: Remember that a function must be continuous at a point to be differentiable there. Also, a sharp corner or a cusp in the graph always indicates non-differentiability.

Question 5. The graph of f is shown below. State with reasons that x values (the numbers), at which f is not differentiable.
Answer:
A function \( f \) is not differentiable at a point \( x_0 \) within its domain if any of the following conditions are met:
(i) The function has a sharp edge (like a 'V' shape) or a sharp peak (like an inverted 'V') at \( x_0 \). At such points, the left-hand derivative and the right-hand derivative are different.
(ii) The function has a vertical tangent line at \( x_0 \). This means the slope becomes infinitely steep.
Looking at the provided graph:
At \( x = -1 \), the graph shows a sharp edge (a 'V' shape). This indicates that the function is not differentiable at \( x = -1 \).
At \( x = 8 \), the graph shows a sharp peak (an inverted 'V' shape). This means the function is not differentiable at \( x = 8 \).
At \( x = 11 \), the graph appears to have a vertical tangent line. A vertical tangent implies an undefined slope, so the function is not differentiable at \( x = 11 \).
Therefore, the given graph is not differentiable at \( x = -1, 8, 11 \).
In simple words: A function cannot be differentiated where its graph has a pointy corner or where the line becomes straight up and down. On this graph, that happens at \( x = -1 \), \( x = 8 \) (sharp corners), and \( x = 11 \) (vertical line).

🎯 Exam Tip: Visually identify points of non-differentiability by looking for sharp corners, cusps, or vertical tangent lines on the graph of the function.

Question 6. If \( f(x) = |x + 100| + x^2 \), test whether \( f'(-100) \) exists.
Answer:
The given function is \( f(x) = |x + 100| + x^2 \). We need to check for differentiability at \( x = -100 \).
First, let's define \( |x + 100| \):
\( |x + 100| = \begin{cases} (x + 100) & ; x + 100 \ge 0 \implies x \ge -100 \\ -(x + 100) & ; x + 100 < 0 \implies x < -100 \end{cases} \)
So, \( f(x) = \begin{cases} -(x + 100) + x^2 & ; x < -100 \\ (x + 100) + x^2 & ; x \ge -100 \end{cases} \)
Now, we evaluate \( f(-100) \). Using the second case for \( x \ge -100 \):
\( f(-100) = (-100 + 100) + (-100)^2 = 0 + 10000 = 10000 \).
Next, let's find the left-hand derivative at \( x = -100 \):
For \( x < -100 \), \( f(x) = -(x + 100) + x^2 \).
\( f'(-100^-) = \lim_{x \to -100^-} \frac{f(x) - f(-100)}{x - (-100)} \)
\( = \lim_{x \to -100^-} \frac{(-(x + 100) + x^2) - 10000}{x + 100} \)
\( = \lim_{x \to -100^-} \frac{-(x + 100) + x^2 - (-100)^2}{x + 100} \)
\( = \lim_{x \to -100^-} \left( \frac{-(x + 100)}{x + 100} + \frac{x^2 - (-100)^2}{x + 100} \right) \)
\( = \lim_{x \to -100^-} \left( -1 + \frac{(x - (-100))(x + (-100))}{x + 100} \right) \)
\( = \lim_{x \to -100^-} (-1 + x - 100) \)
\( = -1 + (-100) - 100 \)
\( = -1 - 100 - 100 = -201 \) -----(1)
Now, let's find the right-hand derivative at \( x = -100 \):
For \( x > -100 \), \( f(x) = (x + 100) + x^2 \).
\( f'(-100^+) = \lim_{x \to -100^+} \frac{f(x) - f(-100)}{x - (-100)} \)
\( = \lim_{x \to -100^+} \frac{((x + 100) + x^2) - 10000}{x + 100} \)
\( = \lim_{x \to -100^+} \frac{(x + 100) + x^2 - (-100)^2}{x + 100} \)
\( = \lim_{x \to -100^+} \left( \frac{(x + 100)}{x + 100} + \frac{x^2 - (-100)^2}{x + 100} \right) \)
\( = \lim_{x \to -100^+} \left( 1 + \frac{(x - (-100))(x + (-100))}{x + 100} \right) \)
\( = \lim_{x \to -100^+} (1 + x - 100) \)
\( = 1 + (-100) - 100 \)
\( = 1 - 100 - 100 = -199 \) -----(2)
From equations (1) and (2), we observe that \( f'(-100^-) \neq f'(-100^+) \) because \( -201 \neq -199 \). Since the left-hand and right-hand derivatives are not equal, \( f'(-100) \) does not exist, meaning the function is not differentiable at \( x = -100 \). The absolute value function creates a sharp point where the derivative is undefined.
In simple words: To see if a function can be smoothly differentiated at a point, we check its slope coming from the left and from the right. Here, at \( x = -100 \), the slopes are different (-201 and -199). This means the function has a sharp point there, so it cannot be smoothly differentiated.

🎯 Exam Tip: When dealing with absolute value functions, always break them into piecewise definitions based on the sign of the expression inside the absolute value. This is crucial for correctly evaluating left-hand and right-hand derivatives.

Question 7. Examine the differentiability of functions in R by drawing the diagrams.
(i) \( |\sin x| \)
Answer:
Let \( f(x) = |\sin x| \). The graph of \( \sin x \) is a wave that goes above and below the x-axis. The absolute value function \( |\sin x| \) takes any negative values of \( \sin x \) and makes them positive, effectively reflecting the parts of the graph below the x-axis upwards. This creates sharp corners wherever \( \sin x = 0 \).
The sine function is zero at \( x = n\pi \), where \( n \) is an integer (\( \dots, -2\pi, -\pi, 0, \pi, 2\pi, \dots \)). At these points, the graph of \( |\sin x| \) will have sharp "V" shapes, meaning the function is not differentiable at these points.
Let's look at some values to understand the shape near these points:
The table below shows how \( |\sin x| \) behaves at key points:

At the points \( x = 0, \pi, 2\pi, 3\pi, \dots \), the graph of \( f(x) = |\sin x| \) has sharp "V" shaped edges. These sharp points occur because the derivative of \( \sin x \) changes sign abruptly (from positive to negative or negative to positive) at these x-intercepts when the absolute value is applied. Therefore, at these points, the left-hand and right-hand derivatives are not equal, making the function not differentiable. The function \( y = |\sin x| \) is not differentiable at \( x = n\pi \) for all integers \( n \in \mathbb{Z} \).
In simple words: The function \( |\sin x| \) creates sharp corners on its graph every time \( \sin x \) crosses the x-axis. These corners happen at \( 0, \pi, 2\pi \) and so on. A function cannot be differentiated at sharp corners, so \( |\sin x| \) is not differentiable at these points.

🎯 Exam Tip: For functions involving absolute values, remember that differentiability often breaks down at the points where the expression inside the absolute value becomes zero, as these points usually correspond to sharp corners or cusps.

(ii) \( |\cos x| \)
Answer:
Let \( f(x) = |\cos x| \). Similar to \( |\sin x| \), the absolute value function \( |\cos x| \) reflects the negative parts of the \( \cos x \) graph upwards, creating sharp corners wherever \( \cos x = 0 \).
The cosine function is zero at \( x = (n + \frac{1}{2})\pi \), which means at \( x = \dots, -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \dots \). At these points, the graph of \( |\cos x| \) will have sharp "V" shapes, indicating that the function is not differentiable there.
Let's examine some values to illustrate this:

At the points where \( \cos x = 0 \) (i.e., \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots \)), the graph of \( f(x) = |\cos x| \) will have sharp corners. This is because the sign of the derivative changes abruptly at these points when the absolute value is applied. For example, at \( x = \frac{\pi}{2} \), the derivative approaches \( -1 \) from the left and \( +1 \) from the right, making them unequal. Hence, the function \( y = |\cos x| \) is not differentiable at \( x = (n + \frac{1}{2})\pi \) for all integers \( n \in \mathbb{Z} \).
In simple words: The function \( |\cos x| \) has sharp corners on its graph every time \( \cos x \) becomes zero. This happens at \( \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \), and so on. Because of these sharp corners, the function cannot be differentiated at these specific points.

🎯 Exam Tip: Functions of the form \( |g(x)| \) are generally not differentiable at points where \( g(x) = 0 \) and \( g'(x) \neq 0 \), because these points create sharp turns in the graph.