Question

for the function f(x)=x^3-48x,
Its local maximum is the Point ( ), ( )
Its Local Minimum is the Point ( ), ( )

Answer 1

f'(x)=3x^2 - 48
f'(x)=0 then 3x^2 - 48=0
x^2-16=0
x=4 or -4
points are (4,4^3 -48);(-4,-4^3 -48)
= (4,16); (-4,-112)
so local maximum is (4,16)
local minimum is (-4,-112)

Answer 2

I see how you got this answer but it should be x^3


for the function f(x)=x^3-48x,
Its local maximum is the Point ( ), ( )
Its Local Minimum is the Point ( ), ( )

Answer 3

he did take the derivative of that

Answer 4

\[f \prime(x)=3x ^{2}-48\]\[3x ^{2}-48=0\]\[3(x ^{2}-16)=0\]\[x ^{2}-16=0\]\[x=\pm4\]Now plug in the values of x and get f(x) for find the point.\[f(4)=4^{3}-48(4)\]\[f(4)=64-192=-128\]\[f(-4)=(-4)^{3}-48(-4)\]\[f(-4)=-64+192=128\]Local Max at (-4,128) and Local Min at (4,-128)

Answer 5

He did not plug the solutions to the derivative back into the original equation.

Answer 6

oh i thought you meant the derivative part... i didn't check all that other stuff

Answer 7

sorry p3ppp......i did a mistake when i was putting f(4) and f(-4)...last part..calculation..