# a. Find the intervals on which f is increasing or decreasing. b. Find the local maximum and...

## Question:

a. Find the intervals on which {eq}f {/eq} is increasing or decreasing.

b. Find the local maximum and minimum values of {eq}f {/eq}.

c. Find the intervals of concavity and the inflection points.

{eq}\displaystyle\; f(x) = x^{4}e^{-x} {/eq}

## Maxima and Minima of a function

A given function has a maxima at the point where the value of the function in its local neighbourhood is maximum and similarly a minima when the value of the function in its local neighbourhood is minimum.

To find maxima or minima or the inflection points we use differentiation.

## Answer and Explanation:

Given a function {eq}\displaystyle f(x) = x^4 e^{-x} {/eq}

**(a)** We need to find the intervals on which {eq}\displaystyle f(x) {/eq} is increasing or decreasing.

For this let's find it's critical numbers.

{eq}\displaystyle f'(x) = 0 = 4x^3 e^{-x} + x^4 e^{-x} (-1) \\ \Rightarrow x^3e^{-x} ( 4 - x ) = 0 {/eq}

So the critical numbers of this function are {eq}\displaystyle x = 0 , 4 {/eq}

We can see that the derivative of the function {eq}\displaystyle f(x) {/eq} is

{eq}\displaystyle f'(x) = x^3 e^{-x} (4-x) < 0 {/eq} for {eq}\displaystyle x < 0 {/eq}

Also, {eq}\displaystyle f'(x) = x^3 e^{-x} (4-x) < 0 {/eq} for {eq}\displaystyle x>4 {/eq}

Just, {eq}\displaystyle f'(x) = x^3 e^{-x} (4-x) > 0 {/eq} for {eq}\displaystyle x \in (0,4) {/eq}

So, the function is **decreasing** in the interval {eq}\displaystyle x \in ( - \infty , 0 ) \cup ( 4 , \infty ) {/eq}

And **increasing** in the interval {eq}\displaystyle x \in (0,4) {/eq}

**(b)** The critical points are {eq}\displaystyle x = \left \{ 0 \ , \ 4 \right \} {/eq}

We can see that the slope of function decreases till x = 0, and then increases. Hence it must be a local minimum.

Also,

We can see that the slope of function increases till x = 4, and then decreases. Hence it must be a local maximum.

c)

{eq}\displaystyle f'(x) = 0 = 4x^3 e^{-x} + x^4 e^{-x} (-1) \\ \Rightarrow x^3e^{-x} ( 4 - x ) = 0 \\ f''(x) = 12x^2 e^{-x} -4x^3 e^{-x} -4x^3 e^{-x}+ x^4 e^{-x} =e^{-x}4x^2(3-x) -x^3e^{-x} ( 4 - x ) =0 {/eq}

It is zero at x=0, thus we can say that x=0 is an inflection point.

For x<0 we can see that f*(x) >0 so the it means f'(x) increases and thus the function is concave up, *

*from x=0 the function becomes concave down or convex.*

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from General Studies Math: Help & Review

Chapter 5 / Lesson 2