Calculus S.M Yusuf Chapter 3 Ex 3.1 Solved Notes & Quiz

Question 1: Verify Rolle's Theorem for the function $f(x) = x^3 - 4x$ on the interval $[-2, 2]$. Find the value(s) of $c$.

Solution:
Step 1: Check Continuity and Differentiability.
Since $f(x) = x^3 - 4x$ is a polynomial function, it is continuous on every closed real interval. Thus, $f(x)$ is continuous on $[-2, 2]$.
Its derivative is $f'(x) = 3x^2 - 4$, which exists for all $x \in (-2, 2)$. Thus, $f(x)$ is differentiable on $(-2, 2)$.

Step 2: Check Endpoint Values.
$$f(-2) = (-2)^3 - 4(-2) = -8 + 8 = 0$$ $$f(2) = (2)^3 - 4(2) = 8 - 8 = 0$$ Since $f(-2) = f(2) = 0$, all three hypotheses of Rolle's Theorem are satisfied.

Step 3: Solve for $c$.
According to Rolle's Theorem, there exists $c \in (-2, 2)$ such that $f'(c) = 0$. $$f'(c) = 3c^2 - 4 = 0$$ $$3c^2 = 4 \implies c^2 = \frac{4}{3} \implies c = \pm \frac{2}{\sqrt{3}}$$ Evaluating numerically: $c \approx \pm 1.155$.
Both values $c_1 = -\frac{2}{\sqrt{3}}$ and $c_2 = \frac{2}{\sqrt{3}}$ belong to the open interval $(-2, 2)$.
Hence, Rolle's Theorem is verified.
Answer: $$c = \pm \frac{2}{\sqrt{3}} \in (-2, 2)$$

Question 2: Verify Rolle's Theorem for $f(x) = (x-1)(x-2)(x-3)$ on $[1, 3]$. Find the corresponding value(s) of $c$.

Solution:
Step 1: Simplify the Function and Check Conditions.
Expanding the function: $$f(x) = (x-1)(x^2 - 5x + 6) = x^3 - 6x^2 + 11x - 6$$ As a polynomial, $f(x)$ is continuous on $[1, 3]$ and differentiable on $(1, 3)$.

Step 2: Evaluate Endpoint Values.
$$f(1) = (1-1)(1-2)(1-3) = 0$$ $$f(3) = (3-1)(3-2)(3-3) = 0$$ Since $f(1) = f(3) = 0$, Rolle's Theorem is applicable.

Step 3: Find $c \in (1, 3)$ where $f'(c) = 0$.
Differentiating $f(x)$: $$f'(x) = 3x^2 - 12x + 11$$ Set $f'(c) = 0$: $$3c^2 - 12c + 11 = 0$$ Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$c = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(11)}}{2(3)} = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = 2 \pm \frac{\sqrt{3}}{3} = 2 \pm \frac{1}{\sqrt{3}}$$ Evaluating numerically: $c_1 = 2 - 0.577 = 1.423$ and $c_2 = 2 + 0.577 = 2.577$.
Both $1.423$ and $2.577$ lie strictly within the open interval $(1, 3)$.
Thus, Rolle's Theorem is fully verified.
Answer: $$c = 2 \pm \frac{1}{\sqrt{3}} \in (1, 3)$$

Question 3: Verify Lagrange's Mean Value Theorem for $f(x) = x^2 - 4x + 3$ on the interval $[1, 4]$. Find $c$.

Solution:
Step 1: Check Continuity and Differentiability.
$f(x) = x^2 - 4x + 3$ is polynomial, so it is continuous on $[1, 4]$ and differentiable on $(1, 4)$.

Step 2: Compute Function Values at Endpoints.
For $a = 1$: $f(1) = (1)^2 - 4(1) + 3 = 1 - 4 + 3 = 0$.
For $b = 4$: $f(4) = (4)^2 - 4(4) + 3 = 16 - 16 + 3 = 3$.

Step 3: Apply the LMVT Formula.
$$\frac{f(b) - f(a)}{b - a} = \frac{f(4) - f(1)}{4 - 1} = \frac{3 - 0}{3} = 1$$ Now calculate $f'(x)$: $$f'(x) = 2x - 4 \implies f'(c) = 2c - 4$$ Set $f'(c) = \frac{f(4) - f(1)}{4 - 1}$: $$2c - 4 = 1 \implies 2c = 5 \implies c = \frac{5}{2} = 2.5$$ Since $2.5 \in (1, 4)$, Lagrange's Mean Value Theorem is verified.
Answer: $$c = \frac{5}{2} \in (1, 4)$$

Question 4: Verify Lagrange's Mean Value Theorem for $f(x) = \sqrt{x}$ on $[1, 4]$. Find $c$.

Solution:
Step 1: Check Hypotheses.
$f(x) = \sqrt{x}$ is defined and continuous on $[1, 4]$.
Its derivative $f'(x) = \frac{1}{2\sqrt{x}}$ exists and is defined for all $x \in (1, 4)$. Hence $f(x)$ is differentiable on $(1, 4)$.

Step 2: Evaluate Endpoints and Secant Slope.
$$f(1) = \sqrt{1} = 1$$ $$f(4) = \sqrt{4} = 2$$ $$\frac{f(4) - f(1)}{4 - 1} = \frac{2 - 1}{3} = \frac{1}{3}$$ Step 3: Solve $f'(c) = \frac{1}{3}$.
$$\frac{1}{2\sqrt{c}} = \frac{1}{3}$$ Cross-multiplying yields: $$2\sqrt{c} = 3 \implies \sqrt{c} = \frac{3}{2} \implies c = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25$$ Since $2.25 \in (1, 4)$, LMVT is successfully verified.
Answer: $$c = \frac{9}{4} \in (1, 4)$$

Question 5: Verify Cauchy's Mean Value Theorem for $f(x) = x^2$ and $g(x) = x^3$ on $[1, 2]$. Find $c$.

Solution:
Step 1: Check Conditions for Cauchy's MVT.
Both $f(x) = x^2$ and $g(x) = x^3$ are continuous on $[1, 2]$ and differentiable on $(1, 2)$.
$g'(x) = 3x^2 \neq 0$ for all $x \in (1, 2)$. Thus all conditions hold.

Step 2: Compute Boundary Values and Derivatives.
$$f(1) = 1^2 = 1, \quad f(2) = 2^2 = 4 \implies f(2) - f(1) = 4 - 1 = 3$$ $$g(1) = 1^3 = 1, \quad g(2) = 2^3 = 8 \implies g(2) - g(1) = 8 - 1 = 7$$ $$f'(x) = 2x \implies f'(c) = 2c$$ $$g'(x) = 3x^2 \implies g'(c) = 3c^2$$ Step 3: Apply Cauchy's MVT Formula.
$$\frac{f'(c)}{g'(c)} = \frac{f(2) - f(1)}{g(2) - g(1)}$$ $$\frac{2c}{3c^2} = \frac{3}{7}$$ Simplifying the left-hand side (since $c \neq 0$ on $(1, 2)$): $$\frac{2}{3c} = \frac{3}{7}$$ Cross-multiplying gives: $$9c = 14 \implies c = \frac{14}{9} \approx 1.556$$ Since $1.556 \in (1, 2)$, Cauchy's Mean Value Theorem is verified.
Answer: $$c = \frac{14}{9} \in (1, 2)$$

Question 6: Verify Cauchy's Mean Value Theorem for $f(x) = \sin x$ and $g(x) = \cos x$ on $\left[0, \frac{\pi}{2}\right]$. Find $c$.

Solution:
Step 1: Verify Hypotheses.
$f(x) = \sin x$ and $g(x) = \cos x$ are continuous on $\left[0, \frac{\pi}{2}\right]$ and differentiable on $\left(0, \frac{\pi}{2}\right)$.
$g'(x) = -\sin x \neq 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.

Step 2: Evaluate Function Values and Derivatives.
$$f(0) = \sin 0 = 0, \quad f\left(\frac{\pi}{2}\right) = \sin \left(\frac{\pi}{2}\right) = 1 \implies f\left(\frac{\pi}{2}\right) - f(0) = 1 - 0 = 1$$ $$g(0) = \cos 0 = 1, \quad g\left(\frac{\pi}{2}\right) = \cos \left(\frac{\pi}{2}\right) = 0 \implies g\left(\frac{\pi}{2}\right) - g(0) = 0 - 1 = -1$$ $$f'(c) = \cos c, \quad g'(c) = -\sin c$$ Step 3: Set up Cauchy's Equation.
$$\frac{f'(c)}{g'(c)} = \frac{f(\pi/2) - f(0)}{g(\pi/2) - g(0)}$$ $$\frac{\cos c}{-\sin c} = \frac{1}{-1}$$ $$-\cot c = -1 \implies \cot c = 1 \implies \tan c = 1$$ Since $c \in \left(0, \frac{\pi}{2}\right)$: $$c = \arctan(1) = \frac{\pi}{4}$$ Since $\frac{\pi}{4} \in \left(0, \frac{\pi}{2}\right)$, Cauchy's MVT is fully verified.
Answer: $$c = \frac{\pi}{4} \in \left(0, \frac{\pi}{2}\right)$$