Question

For following questions check if Rolle's (or) LMVT is applicable or not, if applicable then find c:

(i) $f(x) = x(x + 3)e^{-x/2}$, $x \in [-3,0]$ (ii) $f(x) = \frac{sinx}{e^x}$, $x \in [0, \pi]$ (iii) $f(x) = \frac{x}{x+2}$, $x \in [1,4]$ (iv) $f(x) = \sqrt{x - 1}$, $x \in [2,3]$ (v) $f(x) = 1-x^{2/3}$, $x \in [-1,1]$

Answer 1

No Options

Answer 2

The applicability of Rolle's Theorem or Lagrange's Mean Value Theorem (LMVT) depends on two primary conditions: continuity on the closed interval $[a,b]$ and differentiability on the open interval $(a,b)$. For Rolle's Theorem, an additional condition $f(a) = f(b)$ must be met.

Rolle's Theorem Conditions:

1. $f(x)$ is continuous on $[a, b]$.
2. $f(x)$ is differentiable on $(a, b)$.
3. $f(a) = f(b)$.

If all conditions hold, there exists $c \in (a, b)$ such that $f'(c) = 0$.

Lagrange's Mean Value Theorem (LMVT) Conditions:

1. $f(x)$ is continuous on $[a, b]$.
2. $f(x)$ is differentiable on $(a, b)$.

If both conditions hold, there exists $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

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(i) $f(x) = x(x + 3)e^{-x/2}$, $x \in [-3,0]$

1. Continuity: $f(x)$ is a product of a polynomial and an exponential function, both continuous everywhere. Thus, $f(x)$ is continuous on $[-3,0]$.
2. Differentiability: $f'(x) = e^{-x/2} \left( -\frac{1}{2}x^2 + \frac{1}{2}x + 3 \right)$. This derivative exists for all $x$. Thus, $f(x)$ is differentiable on $(-3,0)$.
3. Check $f(a) = f(b)$: $f(-3) = -3(-3+3)e^{3/2} = 0$. $f(0) = 0(0+3)e^0 = 0$. Since $f(-3) = f(0)$, Rolle's Theorem is applicable.
4. Find c: Set $f'(c) = 0$: $e^{-c/2} \left( -\frac{1}{2}c^2 + \frac{1}{2}c + 3 \right) = 0$ Since $e^{-c/2} \neq 0$, we have $-\frac{1}{2}c^2 + \frac{1}{2}c + 3 = 0 \implies c^2 - c - 6 = 0$. Factoring gives $(c-3)(c+2) = 0$, so $c=3$ or $c=-2$. The value $c=-2$ lies in the interval $(-3,0)$. Applicable: Rolle's Theorem, $c = -2$

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(ii) $f(x) = \frac{\sin x}{e^x}$, $x \in [0, \pi]$

1. Continuity: $f(x) = e^{-x}\sin x$ is a product of continuous functions, so it is continuous on $[0, \pi]$.
2. Differentiability: $f'(x) = e^{-x}(\cos x - \sin x)$. This derivative exists for all $x$. Thus, $f(x)$ is differentiable on $(0, \pi)$.
3. Check $f(a) = f(b)$: $f(0) = \frac{\sin 0}{e^0} = \frac{0}{1} = 0$. $f(\pi) = \frac{\sin \pi}{e^\pi} = \frac{0}{e^\pi} = 0$. Since $f(0) = f(\pi)$, Rolle's Theorem is applicable.
4. Find c: Set $f'(c) = 0$: $e^{-c}(\cos c - \sin c) = 0$ Since $e^{-c} \neq 0$, we have $\cos c - \sin c = 0 \implies \tan c = 1$. For $c \in (0, \pi)$, $c = \frac{\pi}{4}$. Applicable: Rolle's Theorem, $c = \frac{\pi}{4}$

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(iii) $f(x) = \frac{x}{x+2}$, $x \in [1,4]$

1. Continuity: $f(x)$ is a rational function, continuous everywhere except $x=-2$. Since $-2 \notin [1,4]$, $f(x)$ is continuous on $[1,4]$.
2. Differentiability: $f'(x) = \frac{2}{(x+2)^2}$. This derivative exists for all $x \neq -2$. Thus, $f(x)$ is differentiable on $(1,4)$.
3. Check $f(a) = f(b)$: $f(1) = \frac{1}{1+2} = \frac{1}{3}$. $f(4) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$. Since $f(1) \neq f(4)$, Rolle's Theorem is not applicable. However, LMVT is applicable.
4. Find c: Set $f'(c) = \frac{f(b) - f(a)}{b - a}$: $f'(c) = \frac{f(4) - f(1)}{4 - 1} = \frac{\frac{2}{3} - \frac{1}{3}}{3} = \frac{\frac{1}{3}}{3} = \frac{1}{9}$. $\frac{2}{(c+2)^2} = \frac{1}{9} \implies (c+2)^2 = 18 \implies c+2 = \pm\sqrt{18} = \pm 3\sqrt{2}$. $c = -2 \pm 3\sqrt{2}$. $c = -2 + 3\sqrt{2} \approx -2 + 4.242 = 2.242$. This value lies in $(1,4)$. $c = -2 - 3\sqrt{2} \approx -2 - 4.242 = -6.242$. This value is not in $(1,4)$. Applicable: LMVT, $c = -2 + 3\sqrt{2}$

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(iv) $f(x) = \sqrt{x - 1}$, $x \in [2,3]$

1. Continuity: $f(x) = \sqrt{x-1}$ is continuous for $x \ge 1$. Thus, $f(x)$ is continuous on $[2,3]$.
2. Differentiability: $f'(x) = \frac{1}{2\sqrt{x-1}}$. This derivative exists for $x > 1$. Thus, $f(x)$ is differentiable on $(2,3)$.
3. Check $f(a) = f(b)$: $f(2) = \sqrt{2-1} = 1$. $f(3) = \sqrt{3-1} = \sqrt{2}$. Since $f(2) \neq f(3)$, Rolle's Theorem is not applicable. However, LMVT is applicable.
4. Find c: Set $f'(c) = \frac{f(b) - f(a)}{b - a}$: $f'(c) = \frac{f(3) - f(2)}{3 - 2} = \frac{\sqrt{2} - 1}{1} = \sqrt{2} - 1$. $\frac{1}{2\sqrt{c-1}} = \sqrt{2} - 1 \implies 2\sqrt{c-1} = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1$. $\sqrt{c-1} = \frac{\sqrt{2}+1}{2}$. Squaring both sides: $c-1 = \left(\frac{\sqrt{2}+1}{2}\right)^2 = \frac{2+1+2\sqrt{2}}{4} = \frac{3+2\sqrt{2}}{4}$. $c = 1 + \frac{3+2\sqrt{2}}{4} = \frac{4+3+2\sqrt{2}}{4} = \frac{7+2\sqrt{2}}{4}$. This value $c \approx \frac{7+2(1.414)}{4} = \frac{9.828}{4} \approx 2.457$ lies in $(2,3)$. Applicable: LMVT, $c = \frac{7+2\sqrt{2}}{4}$

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(v) $f(x) = 1-x^{2/3}$, $x \in [-1,1]$

1. Continuity: $f(x) = 1 - \sqrt[3]{x^2}$ is defined for all real $x$ and is continuous on $[-1,1]$.
2. Differentiability: $f'(x) = -\frac{2}{3}x^{-1/3} = -\frac{2}{3\sqrt[3]{x}}$. The derivative $f'(x)$ is not defined at $x=0$. Since $0 \in (-1,1)$, $f(x)$ is not differentiable on $(-1,1)$. Applicability: Neither Rolle's Theorem nor LMVT is applicable.