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  • Verify the Rolle’s theorem for each of the functions f(x) = x(x + 3)e-x/2 in [-3, 0].

Verify the Rolle’s theorem for each of the functions
f(x) = x(x + 3)e^{-x/2} in [-3, 0].

Answers (1)

Given: f(x) = x(x + 3)e^{-x/2}

\Rightarrow f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

\Rightarrow f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}

Since, f(x) is multiplication of algebra and exponential function and is defined everywhere in its domain.

\Rightarrow f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}} is continuous at x ∈ [-3,0]

Hence, condition 1 is satisfied.

Condition 2: \Rightarrow f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}

On differentiating f(x) with respect to x, we get,

\\ f^{\prime}(x)=e^{\frac{-x}{2}} \frac{d}{d x}\left(x^{2}+3 x\right)+\left(x^{2}+3 x\right) \frac{d}{d x} e^{-\frac{x}{2}} \text { [by product rule }] \\ \Rightarrow f^{\prime}(x)=e^{-\frac{x}{2}}[2 x+3]+\left(x^{2}+3 x\right) \times\left(-\frac{1}{2} e^{-\frac{x}{2}}\right) \\ \Rightarrow f^{\prime}(x)=2 x e^{-\frac{x}{2}}+3 e^{-\frac{x}{2}}-\frac{x^{2}}{2} e^{-\frac{x}{2}}-\frac{3 x}{2} \mathrm{e}^{-\frac{x}{2}} \\ \Rightarrow f^{\prime}(x)=e^{-\frac{x}{2}}\left[2 x+3-\frac{x^{2}}{2}-\frac{3 x}{2}\right] \\ \Rightarrow f^{\prime}(x)=\frac{e^{-\frac{x}{2}}}{2}\left[x+6-x^{2}\right] \\ \Rightarrow f^{\prime}(x)=-\frac{e^{-\frac{x}{2}}}{2}\left[x^{2}-x-6\right] \\ \Rightarrow \quad(x)=-\frac{e^{-\frac{x}{2}}}{2}\left[x^{2}-3 x+2 x-6\right]

\\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\frac{\mathrm{e}^{-\frac{\mathrm{x}}{2}}}{2}[\mathrm{x}(\mathrm{x}-3)+2(\mathrm{x}-3)] \\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\frac{\mathrm{e}^{-\frac{\mathrm{x}}{2}}}{2}[(\mathrm{x}-3)(\mathrm{x}+2)]

⇒ f(x) is differentiable at [-3,0]

Hence, condition 2 is satisfied.

Condition 3:

\\f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}$ \\$f(-3)=\left[(-3)^{2}+3(-3)\right] e^{\frac{-(-3)}{2}}$ \\$=[9-9] \mathrm{e}^{2 / 2}$ \\$=0$ \\$f(0)=\left[(0)^{2}+3(0)\right] e^{\frac{-0}{2}}$ \\$=0$ \\Hence, $f(-3)=f(0)$ \\Hence, condition 3 is also satisfied. \\Now, let us show that $c \in(0,1)$ such that $f^{\prime}(c)=0$ \\$f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}

$On differentiating above with respect to x , we get $\\ f^{\prime}(x)=-\frac{e^{-\frac{x}{2}}}{2}[(x-3)(x+2)]

\\ Put$ x=c in above equation, we get$\\ \mathrm{f}^{\prime}(\mathrm{c})=-\frac{\mathrm{e}^{-\frac{\mathrm{c}}{2}}}{2}[(\mathrm{c}-3)(\mathrm{c}+2)]

all the three conditions of Rolle’s theorem are satisfied

f’(c) = 0

\begin{aligned} &-\frac{e^{-\frac{c}{2}}}{2}[(c-3)(c+2)]=0\\ &\because-\frac{e^{-\frac{c}{2}}}{2} c a n^{\prime} t \text { be zero }\\ &\Rightarrow(c-3)(c+2)=0\\ &\Rightarrow c-3=0 \text { or } c+2=0\\ &\Rightarrow c=3 \text { or } c=-2\\ &\text { So, value of } c=-2,3\\ &c=-2 \in(-3,0) \text { but } c=3 \in(-3,0)\\ &\therefore c=-2 \end{aligned}

Thus, Rolle’s theorem is verified.

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