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
<h3><u>Explanation</u></h3>
- Given Equation
- Vertex Form
where a-term determines how wide/narrow/upward or downward of the graph.
h-term determines the changes of graph for x-axis
k-term determines the changes of graph for y-axis.
The vertex of parabola is at (h,k).
Therefore, h = 6 and k = 3.
<h3><u>Answer</u></h3><u></u>