Question

Given the function f(x)=x^4+8x^3+18x^2+7 determine the absolute minimum value of f on the closed interval [-4, 1].

Answer 1

Answer: 7

Step-by-step explanation:

To find the absolute minimum we must first take the derivative of the function and find where it equals zero. From there we can use the first derivative test to find where our possible minimum is. We then will test each endpoint and compare for the minimum for this function.

$f(x)=x^4+8x^3+18x^2+7\\f'(x) = 4x^3+24x^2+36x\\\\4x^3+24x^2+36x = 4x(x^2+6x +8)\\\\x = 0\\x = -4\\x = -2\\\\f(-4) = 39\\f(-2) =31\\f(0) = 7\\f(1) = 34$

The lowest value is 7. The absolute minimum of this function is 7. If you were to graph it, we'd see that the graph reflects that as well.