# Find the absolute maximum and absolute minimum values of the function 𝑓(𝑥)=𝑥^3+6𝑥^2−63𝑥+11 over each of the indicated intervals.?

Some one please help. I only got two right.

### 3 Answers

- King LeoLv 78 months agoFavorite Answer
f(x) = x³ + 6x² - 63x + 11

absolute minimum or maximum can occur at stationary points or at endpoints

f(x) = x³ + 6x² - 63x + 11

f'(x) = 3x² + 12x - 63

3x² + 12x - 63 = 0

x² + 4x - 21 = 0

x = -7, x = 3

( a )

endpoints: x = -8 and x = 0

but 3 is outside the interval

f(-8) = 387

f(-7) = 403

f(0) = 11

∴

( 1 ) absolute maximum = 403

( 2 ) absolute minimum = 11

( b )

endpoints: x = -5 and x = 4

but -7 is outside the interval

f(-5) = 351

f(3) = -97

f(4) = -81

∴

( 1 ) absolute maximum = 351

( 2 ) absolute minimum = -97

( c )

endpoints: x = -8 and x = 4

but -7 is outside the interval

f(-8) = 387

f(-7) = 403

f(3) = -97

f(4) = -81

∴

( 1 ) absolute maximum = 403

( 2 ) absolute minimum = -97

- Login to reply the answers

- rotchmLv 78 months ago
Apply the exact same procedure already given to you in:

- Login to reply the answers

- ted sLv 78 months ago
So YOU looked at f ' to see where it was equal to 0....the smallest would yield a local max and the larger a local min....if on a closed interval then you checked the function values at the endpoints...CORRECT ??....if not then DO it now.

- Login to reply the answers