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Q48E

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Found in: Page 224

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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# (a) Use a graph of $f$ to estimate the maximum and minimum values. Then find the exact values. (b) Estimate the value of $x$ at which $f$ increases most rapidly. Then find the exact value.$f\left( x \right) = {x^2}{e^{ - x}}$

(a) The local and local and abs max lies at $$\left( {2,4{e^{ - 2}}} \right)$$, $$\left( {0,0} \right)$$.

(b) The resultant answer is $$x = 2 - \sqrt 2$$.

See the step by step solution

## Step 1: Given data

The given function is $$x{\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {\kern 1pt} f$$.

## Step 2: Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

## (a)Step 3: Sketch the graph

The graph is shown below:

$$\begin{array}{l}x \approx 2\\x \approx 0\end{array}$$

## Step 4: Determine the value of the function

Determine the value of the function $$f(x)$$:

$$\begin{array}{l}f(x) = {x^2}{e^{ - x}}\\f\prime (x) = 2x{e^{ - x}} - {x^2}{e^{ - x}}\\f\prime (x) = x{e^{ - x}}(2 - x)\end{array}$$

Similarly, calculate further:

$$\begin{array}{l}x = 2\\x = 0\end{array}$$

## (b) Step 5: Simplify the expression

The problem of finding exact value of $$x$$ at which $$f$$ increases most rapidly is the same as finding value of $$x$$, such that $${f^\prime }(x)$$ is maximum.

$$f(x) = {x^2}{e^{ - x}}$$, Therefore,$${f^\prime }(x) = 2x{e^{ - x}} - {x^2}{e^{ - x}}$$

To find value of $$x$$, such that $${f^\prime }(x)$$ is maximum, we equate $${f^{\prime \prime }}(x) = 0$$

$${f^{\prime \prime }}(x) = 2{e^{ - x}} - 2x{e^{ - x}} - 2x{e^{ - x}} + {x^2}{e^{ - x}}$$

Simplify further;

$$\begin{array}{l}{f^{\prime \prime }}(x) = \left( {2 - 2x - 2x + {x^2}} \right){e^{ - x}}\\{f^{\prime \prime }}(x) = {e^{ - x}}\left( {{x^2} - 4x + 2} \right)\end{array}$$

to have $${f^{\prime \prime }}(x) = 0$$ is same as to have $${x^2} - 4x + 2 = 0$$, since, $${e^{ - x}}$$ cannot be 0

Therefore, $$x = 2 \pm \sqrt 2 ,x = 2 - \sqrt 2$$ is the answer and not $$x = 2 + \sqrt 2$$, because for the former value of $$x,{f^\prime }(x)$$ is increasing before and decreasing after the point in the neighborhood, therefore a maximum.

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