Given
1. 4 x 1/7
2. 12 x 2 5/6
procedure
1. The answer is 4 x 1/7 is less than 4
2. The answer is 12 x 2 5/6 is greater than 12
What is the relative maximum and minimum of the function?
1 answer:
Take the derivitive
f'(x)=6x^2+2x-11
find where f'(x)=0
f'(x)=0 when x=-1.53089 or x=1.19756
we use a sign chart
test values to see where the signs are
(see attachment)
f'(-2)=(+)
f'(0)=(-)
f'(2)=(+)
max happens when sign changes from (+) to (-)
min happens when sign changes from (-) to (+)
according to the chart, max is at -1.53089 and min is at 1.19756
now evaluate the original function for x=-1.53089 and x=1.19756
f(-1.53089)=12.0078
f(1.19756)=-8.30405
max at (-1.53089,12) and min at (1.19756, -8.30405)
I may have rounded off differently, but
answer is 2nd option
f'(x)=6x^2+2x-11
find where f'(x)=0
f'(x)=0 when x=-1.53089 or x=1.19756
we use a sign chart
test values to see where the signs are
(see attachment)
f'(-2)=(+)
f'(0)=(-)
f'(2)=(+)
max happens when sign changes from (+) to (-)
min happens when sign changes from (-) to (+)
according to the chart, max is at -1.53089 and min is at 1.19756
now evaluate the original function for x=-1.53089 and x=1.19756
f(-1.53089)=12.0078
f(1.19756)=-8.30405
max at (-1.53089,12) and min at (1.19756, -8.30405)
I may have rounded off differently, but
answer is 2nd option
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