Question

Find the local maximum and two local minima of the graph of the following function.

Answer 1

Answer:

max is (3, 1) and min are (2, 0) and (4, 0)

Step-by-step explanation:

Since there is no bounds on this (not a closed interval), the only max and min we can find are local.  The max and min points exist where the first derivative of the function is equal to 0.  That means that we have to find the first derivative.  That is:

$y'=4x^3-36x^2+104x-96$

If you factor this higher-degree polynomial (I used the Rational Root Theorem and then synthetic division), you find that the zeros of the derivative exist at the x values of

2, 3, 4

Therefore, f(2), f(3), and f(4) will either be max values or min values.

f(2) = 0 so the point is (2, 0)

f(3) = 1 so the point is (3, 1)

f(4) = 0 so the point is (4, 0)

As you can see, the max point is (3, 1)

the min points are (2, 0) and (4, 0)