How Derivatives Affect the Shape of a Graph 8

Question

The function 𝑓(𝑥)=4𝑥+4𝑥-1 has one local minimum and one local maximum.

This function has a local maximum at 𝑥=   

with value   

and a local minimum at 𝑥=   

with value   

Answer 1

I'm assuming the function you meant to write was f(x) = 4x³ + 4x² - 1

A local maximum occurs at a peak on the graph and a local minimum occurs at a crest.

At both these points, the derivative is equal to 0.

So we find the derivative of the function and set it equal to 0.

f '(x) = 12x² + 8x = 0

4x(3x + 2) = 0

We see critical values at x = 0 and -2/3.

These represent possible places we have local max and min.

Check intervals at which the derivative is positive and negative by making a number line

<---------- -2/3 ---------- 0 ---------->

The derivative is positive in the right most interval due to the end behavior of a positive polynomial

+

<---------- -2/3 ---------- 0 ---------->

The zero root has a multiplicity of 1 and the -2/3 root also has a multiplicity of one. Because of the odd multiplicities, the sign alternates at 0 and -2/3

+ - +

<---------- -2/3 ---------- 0 ---------->

Where f '(x) is positive, f(x) is increasing and where f '(x) is negative f(x) is decreasing so f(x) should look like this

Inc Dec Inc

<---------- -2/3 ---------- 0 ---------->

At -2/3, our function increases and then decreases, so this is where we have a local max.

At 0, our function decreases and then increases, so this is where we have a local min

Answer 2

𝑓(𝑥)=4𝑥+4𝑥-1 => I guess you meant by 4x^2 + 4x -1.

We are gonna use 1st derivative test

1. find the critical numbers => f'=8x+4 => 8x+4=0 => x=-1/2
2. check the slopes before and after x=-2

=> x=-1, f'(-1) = -8+4 = -4 < 0 /// x=0, f'(0) = 8*(0)+4 = 4>0

1. before x=-1/2 => decreasing, after x=-1/2 => increasing
2. Hence, it is a local minimum: f(-1/2) = 4*1/4+4*(-1/2)-1 = 1-2-1 = -2

Local min -2 at x=-1/2