# Suppose that f(x)=(10-4x)e^x. a. Find all critical numbers of f . b. Use interval notation to...

## Question:

Suppose that {eq}f(x)=(10-4x)e^x {/eq}

a. Find all critical numbers of {eq}f {/eq}.

b. Use interval notation to indicate where {eq}f(x) {/eq} is increasing.

c. Use interval notation to indicate where {eq}f(x) {/eq} is decreasing.

d. List the {eq}x {/eq} coordinates of all local maxima of {eq}f {/eq}.

e. List the {eq}x {/eq} coordinates of all local minima of {eq}f {/eq}.

f. Use interval notation to indicate where {eq}f(x) {/eq} is concave up.

g. Use interval notation to indicate where {eq}f(x) {/eq} is concave down.

## Critical point:

Critical point is understood as:

- a singular point, a point where the derivative does not exist.

- an extreme point of the domain of definition of the function, where the derivative is zero.

Part a.

Find all critical point

{eq}\displaystyle f(x)= (10-4x)e^x\\ \displaystyle f'(x)= \left(-4\right)e^x+e^x\left(10-4x\right)\\ \displaystyle f'(x)= 6e^x-x4e^x\\ \displaystyle f'(x)= 2e^x(3-2x)\\ \displaystyle f'(x)=0 \,\,\, \rightarrow \,\,\,\, 3-2x=0\\ \displaystyle \boxed{ x= \frac {3}{2}} \Rightarrow \textrm {critical points of the function} {/eq}

Part b.c.

To define the sign of the first derivative in each interval, evaluate a point of each interval and verify the sign:

{eq}\left (-\infty , \frac {3}{2}\right ) \,\,\,\, \Rightarrow \,\,\,\, f'(0)=6 > 0 \,\,\,\,\, \textrm { the first derivative is positive in this interval}\\ \left (\frac {3}{2},\infty\right ) \,\,\, \Rightarrow \,\,\,\, f'(2)=- 2e^{-2} < 0 \,\,\,\,\, \textrm { the first derivative is negative in this interval} {/eq}

Therefore,

{eq}\textrm {Decreasing on} \Longrightarrow \boxed{ \left (\frac {3}{2},\infty\right )}\\ \textrm {Increasing on} \Longrightarrow \boxed{ \left (-\infty , \frac {3}{2}\right )}\\ {/eq}

Part d.e.

To the left of the {eq}x=\frac {3}{2} {/eq} the function is increasing and to the right of the point the function is decreasing, we can conclude that for {eq}x=\frac {3}{2} {/eq} the function reaches a relative maximum.

Part f.g.

To determine the function's concavity, we find and analyze the second derivative:

{eq}\displaystyle f'(x)= 6e^x-x4e^x\\ \displaystyle f''(x) = 6e^x-4e^x-4e^xx\\ \displaystyle f''(x) = 2e^x-x4e^x\\ \displaystyle f''(x) = 2e^x(1-2x)\\ \displaystyle f''(x) = 0 \,\,\,\, \rightarrow 1-2x =0 \\ \boxed{x= \frac{1}{2}} \,\, \Rightarrow \,\,\,\, \textrm {Points where the second derivative is zero} {/eq}

To define the sign of the second derivative in each interval, evaluate a point of each interval and verify the sign:

{eq}\left (-\infty, \frac{1}{2}\right) \,\,\,\, \rightarrow \,\,\,\, f''(0)=2 > 0 \,\,\,\,\, \textrm { the second derivative is positive in this interval }\\ \left (\frac{1}{2},\infty\right) \,\,\,\, \rightarrow \,\,\,\, f''(1)= -2e^{-1}< 0 \,\,\,\,\, \textrm { the second derivative is negative in this interval } {/eq}

Therefore,

{eq}\textrm {Concave down on} \Longrightarrow \boxed{ \left (\frac{1}{2},\infty\right)}\\ \textrm {Concave up on} \Longrightarrow \boxed{ \left (-\infty, \frac{1}{2}\right) }\\ {/eq}