Question

The derivative of a function is g'(x)=3x^2-4x+1 and passes through (2,3). How do I find g(x)?

Answer 1

\[\large g'(x)=3x^2-4x+1\]They also told us that,\[\large g(2)=3\]We can use this piece of information later.

To find g(x), we need to anti-differentiate. Are you familiar with this idea? :D

Answer 2

Somewhat - you do the reverse of what you do to find the derivative, correct?

Answer 3

Yes.
And since we're only dealing with powers of x, it won't be too difficult.

We'll apply the "Power Rule for Integration".
It's basically the opposite of the Power Rule of Derivatives, but also in the reverse order.

For derivatives, we multiply by the power, THEN lower the power by one.
For anti, we raise the power by one, THEN divide by the new power.

Example:\[\huge x^n\]The anti-derivative will give us,\[\huge \frac{x^{n+1}}{n+1}\]

Answer 4

Here is how the first term would change,
\[\large 3x^2 \qquad \rightarrow \qquad \frac{3x^{2+1}}{2+1} \qquad \rightarrow \qquad x^3\]
Any confusion? :D

Answer 5

So once I anti-differentiate g'(x), I should have my answer?

Answer 6

The only thing we have to be careful about is - when you anti-differentiate an unknown CONSTANT will pop back into the problem.
Why?
Because if g(x) was some function which included let's say... +4 at the end of it.
Then the derivative would end up giving us,\[\large g'(x)=3x^2-4x+1+0\]See how the constant disappeared?
We need to find out if there is a constant in the original function. So we throw a +C onto the end of our new found g(x).

And we use the "Initial Condition", the information they gave us about g(x), to find C.

Answer 7

So if you did your anti-differentiation correctly, you should have something like,\[\large g(x)=x^3-2x^2+x+C\]

We'll use the g(2)=3 Information to find C.\[\large 3=2^3-2\cdot2^2+2+C\]

Answer 8

Anti differentiation is called integration.