What are the absolute extrema of # f(x)= 6x^3 − 9x^2 − 36x + 3 in [-4,8]#?

Question

What are the absolute extrema of # f(x)= 6x^3 − 9x^2 − 36x + 3  in [-4,8]#?

Aurora Alvarado · Accepted Answer

# (-4,-381) # and # (8,2211) #

In order to find the extrema, you need to take the derivative of the function and find the roots of the derivative.

i.e. solve for # d/dx [f(x)] = 0 # , use power rule:

#d/dx [6x^3 - 9x^2-36x+3 ] = 18x^2-18x-36 #

solve for the roots: # 18x^2-18x-36 = 0 # # x^2-x-2 = 0 # , factor the quadratic: # (x-1)(x+2) = 0 # # x = 1, x = -2 #

# f(-1) = -6-9+36+3 = 24 # #f(2) = 48-36-72+3 = -57 #

Check the bounds: # f(-4) = -381 # # f(8) = 2211 #

Thus the absolute extrema are # (-4,-381) # and # (8,2211) #

Charles Arocha · Answer

undefined To find the absolute extrema of $ f(x) = 6x^3 - 9x^2 - 36x + 3 $ in the interval $[-4, 8]$, you need to evaluate the function at the critical points and endpoints within that interval.

1. Find the critical points by taking the derivative of $ f(x) $, setting it equal to zero, and solving for $ x $.
2. Evaluate $ f(x) $ at the critical points and at the endpoints of the interval.
3. The largest and smallest values obtained in step 2 will be the absolute maximum and minimum, respectively.

Derivative of $ f(x) $:
$$ f'(x) = 18x^2 - 18x - 36 $$

Setting $ f'(x) = 0 $ and solving for $ x $:
$$ 18x^2 - 18x - 36 = 0 $$
$$ x^2 - x - 2 = 0 $$
$$ (x - 2)(x + 1) = 0 $$

So, critical points are $ x = 2 $ and $ x = -1 $.

Evaluate $ f(x) $ at critical points and endpoints:
$$ f(-4) = 6(-4)^3 - 9(-4)^2 - 36(-4) + 3 = 339 $$
$$ f(2) = 6(2)^3 - 9(2)^2 - 36(2) + 3 = -123 $$
$$ f(8) = 6(8)^3 - 9(8)^2 - 36(8) + 3 = 1479 $$
$$ f(-1) = 6(-1)^3 - 9(-1)^2 - 36(-1) + 3 = 48 $$

The absolute maximum is $ f(8) = 1479 $ and the absolute minimum is $ f(2) = -123 $.