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lol. It took me time to realize that "critical numbers" just means "critical points". :P

Do you know how to find y'?

f'(x)= 9x^2+24x+5
correct??

close

(15x)'=15

oh okay now what do i do after this ?
f'(x)= 9x^2+24x+15

Please refer to my instructions.
Set f'(x)=0 and solve for x :)

i cant seem to be able to solve for x
i got
3x(3x+8)=15

This is a quadratic equation
You may use the quadratic formula

yeah i did that and i got 8 and -8 ; but the answers in the book are -1 and -5/3

So you have 9x^2+24x+15=0
correct?

\[ax^2+bx+c=0 => x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
Did you use this formula?

yes i did

Then you need to try again, because 8 and -8 is not what you should get.

What is a=?
What is b=?
What is c=?

a=9 b=24 c=15
correct??

\[x=\frac{-24 \pm \sqrt{24^2-4(9)(15)}}{2(9)}\]

Use order of operations to evaluate what is inside the radical