Question

Help Please! Been suck for an hour now...
Find the function with the given derivative whose graph passes though the point P.
g'(x)=(1/x^2)+2x , P(-1,1)

Answer 1

integrate the function ... you might know it as antiderivative?

Answer 2

\[\int f(x)~dx\to F(x)+C\]

Answer 3

How do you anitderive the function?

Answer 4

as far as functions go; this is a pretty tame function. What do you know about an antiderivative? or derivatives for that matter?

Answer 5

i know that an anitderivative is a function of f(x) whose derivative passes though a given point?

Answer 6

you should have gone over some rules of derivatives ... i see one rule that would be most beneficial here. The power rule.\[\frac{d}{dx}[x^n]=n~x^{n-1}\]

Answer 7

Oh.. I aware of the rules, but I have a hard time applying them...

Answer 8

I'm*

Answer 9

the antiderivative of this is working it backwards ....
\[\int n~x^{n-1}~dx\to\frac{n}{n-1+1}x^{n-1+1}+C\]\[\frac{n}{n}x^{n}+C\]\[x^{n}+C\]

Answer 10

add one to the exponent and divide it out ...

Answer 11

I haven't learned integrals yet.. is there a way you could show it without the

Answer 12

\[\frac{1}{x^2}+2x \]
\[x^{-2}+2x \]
\[\frac{1}{-2+1}x^{-2+1}+\frac{1}{1+1}2x^{1+1}+C \]

Answer 13

there is ot symbol for "antiderivative", so not really ...

Answer 14

okay. but why are powers adding? dont they minus?

Answer 15

you are UNDOING a derivative ... so no, they dont minus.

In order to undo multiplication, we divide. In order to undo subtraction, we add.

Answer 16

ohhh i see. okay.

Answer 17

\[\frac d{dx}[x^2]\to2x^{2-1}\]
\[\int 2x^{2-1}~dx\to\frac{2}{2-1+1}x^{2-1+1}+C=x^2+C\]

Answer 18

wow, that was a huge step.. haha okay.

Answer 19

when we do this "undoing" power rule (an inverse operation) on your content, we get:
\[\int~x^{-2}+2x~dx\to~\frac{1}{-2+1}x^{-2+1}+\frac{1}{1+1}2x^{1+1}+C\]
\[-x^{-1}+x^2+C\]

Answer 20

you are given a point in order to find a suitable value for C

P(-1,1)
\[1 = (-1)^{-1}+(-1)^2+C\]
\[1 = -1+1+C\]
\[1 = C\]

Answer 21

so the function is \[-x^-1+x^2+C\]

Answer 22

and we evaluated it at the point (-1,1) X=-1 which was just plugged into the function?

Answer 23

@amister64

Answer 24

yes
\[g'(x) \to G(x)+C\]then evaluate it at the given point to determine the appropriate value of C

Answer 25

and of course you have to read thru my typos :) i dropped a negative on accident when I solved for C

Answer 26

I believe it was correct though right? C=1? what negative are you talking about that you dropped?

Answer 27

\[that~one \to\color{red}{(-)}x^{-1}+x^2+C\]
\[1=-(-1)^{-1}+(-1)^2+C\]
\[1=1+1+C\]
\[-1=C\]

Answer 28

Ohhhhh!! okay. I was looking for your mistake but i couldn't find it haha. so C=-1!

Answer 29

:)

Answer 30

Thank you so much! I really appreciate all the help! (:

Answer 31

youre welcome, and good luck ;)