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Essential Calculus: Early Transcendentals
Found in: Page 224
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

(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 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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