Question

Given f′′(x)=2x+4 and f′(−3)=0 and f(−3)=−1.

find f'(x) and f(3)

Hint: start with integrating the second derivative function, and use the "fixed point" f′(−3)=0. Then solve for C, giving you the first derivative function. Now integrate this function, and use the second fixed point given to solve the second part of the problem.

Answer 1

\[f''(x)=2x+4\]
\[f'(x)=x^2+4x+C\]

Answer 2

you know that
\[f'(-3)=0\] so you know
\[(-3)^2+4\times -3+C=0\]
\[9-12+C=0\]
\[C=3\]
so
\[f'(x)=x^2+4x+3\]

Answer 3

now repeat the process for the next one

Answer 4

\[f(x)=\frac{1}{3}x^3+2x^2+3x+C\] and put
\[f(-3)=-1=\frac{1}{3}(-3)^3+2(-3)^2+3(-3)+C\] and solve for C

Answer 5

not getting it. I get 5 for C, then what?

Answer 6

then i guess you are done right?

Answer 7

\[-9+18-9+C=-1\]
\[C=-1\] is what i got, n ot 5

Answer 8

then i guess you are done right?

Answer 9

I'm confused. Why solve for f(-3) when it's asking for f(3)?

Answer 10

ok i did not solve for f(-3)

Answer 11

it says
and f(−3)=−1.
and i used that to solve for C. i got C = -1

Answer 12

so how would I get f(3) then? nothing that I'm doing seems to work. Last thing I tried to input was ((x^3+6x^2)/3)-1

Answer 13

ok i think we have
\[f(x)=\frac{1}{3}x^3+2x^2++3x-1\] yes?

Answer 14

sorry i mean
\[f(x)=\frac{1}{3}x^3+2x^2+3x-1\]

Answer 15

unless i made a mistake somewhere with the constants. now to find
\[f(x)=\frac{1}{3}3^3+2\times 3^2+3\times 3-1\] i get
\[9+18+9-1=35\]

Answer 16

ok, that worked. Now I'm gonna take some time to see where I went wrong. Thanks