Question

find the function f(x) that has the given derivative and the given value.
a) f '(x)=5x^2+3x-2 and f(1)=-1
b) f '(x)= sin(x)+2sec(x)tan(x) and f(pi)=1

Answer 1

So to get from f'(x) to f(x), we need to anti-differentiate.

We have a nice handy rule for dealing with powers of x.
There is a Power Rule for anti-differentiation.

It's the same as the Power Rule for derivatives, except in reverse, and in the reverse order.

We just need to remember that a constant will pop up every time we apply anti-differentiation. It's more formally called "Integration", but if you're just getting an introduction to this type of procedure then you're probably more familiar with the other terminology. :3

Answer 2

Are you familiar with how to apply this rule?
Or we need to review? :D

Answer 3

i need a review

Answer 4

i have to find c first right?

Answer 5

So when we differentiated, we brought the old power down, then decreased the power by 1 to get our new power.

For anti differentiation, we'll raise the power by 1, then divide by the new power. And of course a constant will show up :)
\[\huge x^n \qquad \rightarrow \qquad \frac{x^{n+1}}{n+1} + c\]

Answer 6

Here is a quick example :)
\[\huge x^3 \qquad \rightarrow \qquad \frac{x^4}{4} \qquad \rightarrow \qquad \frac{1}{4}x^4\]Sometimes you'll prefer to write it like this, with the division as a fraction. So it's good to get used to both notations.

Answer 7

ok it wil be 5x^3/3 +3x^2/2 - 2x

Answer 8

+c

Yes ok looks good so far! :)

Answer 9

From here, we'll use the initial-condition to find the missing c value.

Answer 10

so is my c going to be -1 for the first a)

Answer 11

\[\large f(x)=\frac{5}{3}x^3+\frac{3}{2}x^2-2x+c\]
\[\large f(1)=-1 \qquad \rightarrow \qquad -1=\frac{5}{3}1^3+\frac{3}{2}1^2-2\cdot 1+c\]

Answer 12

Understand what we did there? We plugged in x=1, setting it equal to -1,based on our initial-condition.
And we can now solve for c.

I think I got something else besides -1 for c, unless I made a mistake.

Answer 13

ok i m getting -2.166 for c is that you getting?

Answer 14

it looks not right to me

Answer 15

\[\large -1-\frac{5}{3}-\frac{3}{2}+2=c\]\[\large 1-\frac{10-9}{6} \qquad = \qquad 1-\frac{1}{6} \qquad = \qquad \frac{5}{6}\]
Mmmm this is what I'm coming up with, lemme check my work again though... It's easy to make mistakes with fractions :)

Answer 16

Yah I think that's right.
Make sure you leave it as a fraction, not a decimal. Tsk tsk! :o

Answer 17

ok thank you

Answer 18

Then the final step would be to write your equation with the new C value that you've found. :)
\[\large f(x)=\frac{5}{3}x^3+\frac{3}{2}x^2-2x+\frac{5}{6}\]