Question

What is the maximum and minimum of the function? f(x)=x^3 + 3x^2 – 24x and v(x)=(75-2x)(46-2x)x

Answer 1

Are these two separate problems?

Answer 2

yes

Answer 3

You want to find the derivative of the equation, and set that equal to 0 to find the maxes and mins

Answer 4

Any values for x that satisfy that are called critical points. Critical points can be either maxes or mins. That is dependent on the value of the derivative at points directly adjacent to those critical points. If the left side is negative and the right side is positive, then it is a minimum (Imagine an arc where it first drops--negative derivative value-- and then rises--positive derivative value)

Answer 5

i dont know how to solve it

Answer 6

Derivative--->Set equal to 0--->Solve for x values--->Find values of derivative at adjacent points to those x values

Answer 7

f'(x)=3x^2 + 6x – 24
3(x-2)(x+4)
i dont knw what to do next

Answer 8

Set all that equal to 0.

Answer 9

Then solve for what possible x values make that statement true.

Answer 10

x=2;x=-4
can u show me what to do next because i dont have idea what to do please

Answer 11

Those are the critical points, so most of the work is done. Now you just need to know if they are maxes or mins. To do this, plug in values that are to the left (smaller) or right (greater) of each. For example, you have x=2 for one critical point, you can use 1 and 3 as the adjacent points.