Question

Can someone tell me how to solve this?

Answer 1

ok

Answer 2

\[x=3\div2 \pm \sqrt{17\div4}\]

Answer 3

\[x=\frac{ 3 }{ 2 }\pm \sqrt{\frac{ 17 }{ 4 }}\]

Answer 4

It started as:
Use the second derivative test to classify the local (relative) extrema of
\[f(x)=2x ^{3} -9x ^{2}+12x+2\]
I got it down to what I posted above, but I dont know how to solve from there to get x=1,2. Then I got to derive f(x) twice and plug in 1 and 2, but all I dont know how to do is get 1 and 2 from the equation in th elast post. TY

Answer 5

12x-18=0

Answer 6

x=18/12

Answer 7

12x-18=0??? What happened to 12x?

Answer 8

oh wait not 12x 2x^3

Answer 9

I know how to derive it twice, but not how to solve the thing in my first post.

Answer 10

the above one x=3/2...... this the value of x you can't do anything.... you could rewrite it.

but how you end up with it.....

Answer 11

Yes that one, x=3/2... IDK how tpo get x=1,2. I'm studying for my exam tomorrow in calculus, and the answer sheet says x=1,2, and they show everystep up to x=3/2...., but not how they got x to = 1 and 2. I was wondering if you could explain that...

Answer 12

okay

Answer 13

you just need first D

Answer 14

6x^2-18x+12=0

Answer 15

divid by 6
x^2-3x+2

Answer 16

(x-1)(x-2)=0

Answer 17

so x=1 or x=2

Answer 18

wow, that was so much easier then I made it to be...

Answer 19

Thanks soo much!!!

Answer 20

you are welcome