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?
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Answer/Step-by-step explanation:
Note:
<u>Graphing From An Equation: </u>
Equation are frequently written in slope-intercept form { y = mx+b}.
Which "m" represents the slope and "b" represents the y-intercept.
<u>Solutions and Equation:</u>
Solutions to linear equations contain any points located on the graphed line.
Ordered Pair = Solution to an equation if it's values are substituted in the equation which makes it true
Solve:
{x = 0 } 3(0) - 4y = 12
-4y = 12 (0, -3 )
-4y/-4 = 12/-4
y = -3
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
{y = 0} 3x-4(0) =12
3x = 12 (4,0)
3x/3 = 12/3
x = 4
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<u><em>~Lenvy~</em></u>
