1. ## A cubic Polynomial

Hi,
There is a problem:
For f(x)=(x^3)+(6x^2)-15x+k
The absolute minimum and absolute maximum values on the interval [-10,2] have the same absolute value. Find the value of k.

2. Originally Posted by spred
Hi,
There is a problem:
For f(x)=(x^3)+(6x^2)-15x+k
The absolute minimum and absolute maximum values on the interval [-10,2] have the same absolute value. Find the value of k.
Absolute minimums and maximums occur where the derivative is 0.

So

$\displaystyle f'(x) = 3x^2 + 12x - 15$

$\displaystyle 0 = 3x^2 + 12x - 15$

$\displaystyle 0 = x^2 + 4x - 5$

$\displaystyle 0 = (x + 5)(x - 1)$

$\displaystyle x = -5$ or $\displaystyle x = 1$.

This means the critical values are $\displaystyle f(-5)$ and $\displaystyle f(1)$.

Since your minimum and maximum have the same absolute value,

$\displaystyle |f(-5)| = |f(1)|$

$\displaystyle |(-5)^3 + 6(-5)^2 -15(-5) + k| = |1^3 + 6(1)^2 - 15(1) + k|$

$\displaystyle |-125 + 150 - 75 + k| = |1 + 6 - 15 + k|$

$\displaystyle |k - 50| = |k - 8|$

$\displaystyle (k - 50)^2 = (k - 8)^2$

$\displaystyle k^2 - 100k + 250 = k^2 - 16k + 64$

$\displaystyle 186 = 84k$

$\displaystyle k = \frac{186}{84}$

$\displaystyle k = \frac{31}{14}$.