What is the relative maximum and minimum of the function?
1 answer:
Take the derivitive
f'(x)=3x^2+12x
find where it equals zero
it equal zero at x=0 and x=-4
find the y values
f(0)=-36
f(-4)=-4
the critical points are (0,-36) and (-4,-4)
make sign chart
evalutat f'(x) at x=-5 and x=-1 and x=1, and see their signs
f'(-5)=(+)
f'(-1)=(-)
f'(1)=(+)
see below attachment
max is where sign changes from (+) to (-)
min is where sign changes from (-) to (+)
so
max at (-4,-4)
min at (0,-36)
none of the options are correct, do you have te right problem?
f'(x)=3x^2+12x
find where it equals zero
it equal zero at x=0 and x=-4
find the y values
f(0)=-36
f(-4)=-4
the critical points are (0,-36) and (-4,-4)
make sign chart
evalutat f'(x) at x=-5 and x=-1 and x=1, and see their signs
f'(-5)=(+)
f'(-1)=(-)
f'(1)=(+)
see below attachment
max is where sign changes from (+) to (-)
min is where sign changes from (-) to (+)
so
max at (-4,-4)
min at (0,-36)
none of the options are correct, do you have te right problem?
You might be interested in
Answer:
It would be between the 0 and the 1 on the horizontal line
0.5 is where the blue dot is, then for 0.5 , 4 just go up 4 from there
