Suppose that f(x) = (x + 1)(x − 1)^2.

Question

a)Find all critical values of f(x).

 (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

 

(b)

Find the x-coordinates of all local maxima of f(x).

 (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

x = 

 

(c)

Find the x-coordinates of all local minima of f(x).

 (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

x = 

 

Answer 1

In order to find the critical values we need to take the first derivative and set it equal to 0.

We have

ƒ(x) = (x+1)(x-1)²

ƒ(x) = (x+1)'∗(x-1)² + (x+1)∗((x-1)²)'

ƒ'(x) = (x-1)² + (x+1)∗2∗(x-1)

ƒ'(x) = x²-2x+1+2x²-2x+2x-2

ƒ'(x) = 3x²-2x-1

So now that we have the first derivative let's set it equal to 0.

We get

3x²-2x-1 = 0

I will be using the quadratic formula to solve for x.

Remember that the Quadratic Formula is: x = (-b±√(b2-4ac))/2a

So,

x = (2±√(4+12))/6

x = (2±√(16))/6

x = (2±4)/6

x = (1±2)/3

x = 1, -1/3

Hence, are critical values are x = 1, -1/3.

Now to find the local extrema let's take the second derivative and perform the Second Derivative Test.

Recall ƒ'(x) = 3x²-2x-1.

So the second derivative would be,

ƒ''(x) = 6x-2.

Now we plug in the critical values and check wether they are less than or greater than 0.

Let's test x = 1,

f''(1) = 6(1)-2 = 6-2 = 4.

We know 4 > 0, so by the Second Derivative Test x = 1 is a local minima.

Now let's plug in x = -1/3,

f''(-1/3) = 6(-1/3) - 2 = -2 - 2 = -4

We know -4 < 0, so by the Second Derivative Test x = -1/3 is a local maxima.

So the final answers are,

Critical Values: x = 1, -1/3

Local Maxima: x = -1/3

Local Minima: x = 1

I hope this helps!