Question

Find the function with the given derivative whose graph passes through the point p g'(x)= `1/x^2`+2x ,p(-1,1) HELP ME PLEASE??

Answer 1

`g(x) = -x^-1+x^2-1`

Explanation:

We have

`int g'(x)dx = g(x) = int (x^-2+2x)dx = -x^-1+x^2+C_0` but

`g(-1) = 1 rArr -(-1)^-1+(-1)^2+C_0 = 1` and then

`C_0 = -1` so the sought function is

`g(x) = -x^-1+x^2-1`

Answer 2

See below.

Explanation:

If g'(x)=1/x^2+2x, then `g(x)=int(1/x^2+2x) dx`

`int(x^(-2)+2x)dx=-x^(-1)+x^2+c=-1/(x)+x^2+c`

`g(x)=-1/x+x^2+c`

If `g(x)` passes through point P `(-1,1)` then:

`1=-1/(-1)+(-1)^2+c`

`1=1+1+c=>c=-1`

`g(x)=-1/x+x^2-1`

Answer 3

`g(x)=x^2-1/x-1`

Explanation:

I'll do it without integral usage (in case you need it that way)

Graph `C_g` passes trough `P(-1,1)` so that means `g(-1)=1`

We have `g'(x)=1/x^2+2x` ,

`(g(x))' = (-1/x)' +2*(x^2/2)'`

`(g(x))'=(x^2-1/x)'`
`g,x^2-1/x` continuous and differentiable
and `(g(x))'=(x^2-1/x)'` so there is one `c` `in` `RR` for which

`g(x)=x^2-1/x+c`

• `g(-1)=1` `<=> c+2=1` `<=> c=-1`

Eventually `g(x)=x^2-1/x-1`