Challenge: If f'(x)=6x^2+3 and f(1)=8 What is f(x)?

right

Do you know how to do this

Is it a challenge? Or just something youwould like assistance with? c:

no it is a challenge

What I am meaning is that is what the problem says and I don't know how to solve it

oh lol

So to go from \(\large f'\) to \(\large f\) you have to anti-differentiate the function. Do you know how to find the anti derivative of \(\large f'(x)\)?

no

Hmm, no? Haven't you learned that yet? You will need that understanding to do this problem. Have you heard of `Integration` maybe? :) You're usually introduced to this topic with the term anti-differentiation, but maybe you've already skipped ahead to integrals.

wait do you mean this f(x)={6x^2+3dx {6x^2dx+{3dx (6/3x^3)+3x+c 2x^3+3x+c c=-2x^3-3x c=-2(1)^3-3(1) c=-2

Yes the process looks good. I'm not quite sure about your \(\large c\) though.

ok so what do I do from here I am stuck

\[\large f'(x)=6x^2+3\]\[\large f(x)=2x^3+3x+c\] They told us that,\[\large f(1)=8\]So let's plug this in,\[\large 8=2(1)^3+3(1)+c\]

c=10

Hmm

I'm not sure where you're getting 10 from :O I'm coming up with c=3

I took 8=2(1)^3+3(1)+c 8=2+c 8-2=c c=6

8=2+3+c

oh ok I gotcha c=3 so then that is my answer for f(x)=3?

\[\large \color{orangered}{c=3}\]\[\large f(x)=2x^3+3x+\color{orangered}{c}\] Plug in your c value to establish your final answer for f(x).

oh ok so the final answer f(x)=2x^3+3x+3

yay good job!

ok thanks

Don't label your problem, "A Challenge". That generally describes some sort of Math Puzzle or Difficult Problem that you've stumbled across. I think that's what the fuss was about at the start of the thread. Just call it a homework problem or something c: heh

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