Question

verify rolle's theorem for the function f(x) = x(2-x)e^3x/4 in [0,2]. please give the full answer with step by step explanation..​

Answer 1

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Answer 2

Given:

$f(x)=x(2-x)e^{\frac{3x}{4} }$

To prove:

The Rolle's theorem on the given function.

Explanation:

For a given function $f(x)$, Rolle's theorem is applicable only if,

1. The function is continuous on the closed interval $[0,2]$.
2. The function is differentiable on the open interval $(0,2)$ such that $f(a)=f(b)$.

If all the conditions are proved satisfactorily for a function then, according to Rolle's theorem, in the interval $(a,b)$ there will be some value of $c$ in between the interval $a$ and $b$ for which the function $f'(c)=0$

Solution:

According to the question, we have the given function as

$f(x)=x(2-x)e^{\frac{3x}{4} }$  for the interval $[0,2]$.

Step 1

If we take, $f(0)$ for the given function, we get

$f(0)=(0)(2-(0))e^{\frac{3(0)}{4} }$

$f(0)=0$

And, for $f(2)$, we get

$f(2)=(2)(2-(2))e^{\frac{3(2)}{4} }$

$f(2)=0$

$f(1)=f(2)=0$

Since the function gives 0 for the whole interval hence, the function is continuous.

Step 2

The given function is differentiable on the closed interval.

Hence,

We can apply Rolle's theorem.

Let's consider a point $c$ in the interval $[0,2]$  where $f'(x)=0$.

$f'(x)=(2-2x)e^{\frac{3x}{4} }+e^{\frac{3x}{4} } \frac{3}{4} (2x-x^{2} )$

$f'(x)=e^{\frac{3x}{4} } [(2-2x)+\frac{3}{4}(2x-x^{2} ) ]$

$f'(x)=e^{\frac{3x}{4} } [\frac{8-8x+6x-3x^{2} }{4} ]$

$f'(x)=e^{\frac{3x}{4} } [\frac{-3x^{2} -2x+8}{4} ]$

Now, we have, for $c$ , $f'(x)=0$

$e^{\frac{3x}{4} } [\frac{-3x^{2} -2x+8}{4} ]=0$

$-3x^{2} -2x+8=0$

$3x^{2} +2x-8=0$

$3x^{2} +6x-4x-8=0$

$3x(x+2)-4(x+2)=0$

$x=\frac{4}{3} ,-2$

Since, $-2$ ∉ $[0,2]$ .

Hence, $c=\frac{4}{3}$

Final answer:

Hence, the value of $c=\frac{4}{3}$.