if x = -3 is a critical point of y = 4x³ + 12x² - 36x -170 determine whether it is a local minimum, local maximum, or neither?

Question

I need to know how to work this

Answer 1

Hello Hollie.

 

A critical point is a point at which the first derivative is either 0 or does not exist. Generally, if the second derivative is negative at some point, the curve is concave down at that point and if the second derivative is positive at some point, the curve is concave up at that point. If a critical point is a local minimum, then the curve must be concave up at that point, and if a local maximum, then concave down.

 

y = 4x^3+12x^2-36x-170. Then the first derivative is:

dy/dx = 12x^2+24x-36. dy/dx(-3) = 12*9-24*3-36 = 0 as expected.

The second erivative is:

d²y/dx² = 24x+24, and d²y/dx²(-3) = 24*(-3)+24 = -48 < 0.

 

So x=-3 is a local maximum.

 

If you have a graphing calculator, you should try plotting the curve and checking what the graph looks like at x=-3

 

Answer 2

Determine the value of the polynomial at x = -4 and x = -2. If they value of the polynomial is greater at both of them then the polynomial has a minimum at x = -3. Conversely, if the value of the polynomials is lesser at both of them then

the polynomial has a maximum at x = -3

 

f(-4) < f(-3) < f(-2), f(-3) is a local maximum

f(-4) > f(-3) < f(-2), f(-3) is a local minimum