# A cubic Polynomial

• Nov 4th 2009, 11:30 AM
spred
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.
• Nov 4th 2009, 12:22 PM
Prove It
Quote:

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

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

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

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

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

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

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

Since your minimum and maximum have the same absolute value,

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

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

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

$|k - 50| = |k - 8|$

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

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

$186 = 84k$

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

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