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OpenStudy (anonymous):

equation of the tangent line to f(x)3x+1 at x=1

OpenStudy (solomonzelman):

\(\large\color{black}{ \displaystyle f(x)=3x+1 }\)

OpenStudy (anonymous):

Yes

OpenStudy (solomonzelman):

interesting you are finding a tangent line to tangent line ?

OpenStudy (anonymous):

so it would just be 3 correct

OpenStudy (solomonzelman):

I will disappoint you that it will be this same line.

OpenStudy (solomonzelman):

f`(x)=3

OpenStudy (solomonzelman):

yes, but finding tangent line to a line is silly.

OpenStudy (solomonzelman):

it is going to be this very same line 3x+1

OpenStudy (solomonzelman):

maybe you have miss-written the function f(x) ?

OpenStudy (anonymous):

this is what the whole problem says. " Find the equation of the tangent line to f(x)=3x+1 when x=1"

OpenStudy (solomonzelman):

I will show you the abstract side of it.

OpenStudy (anonymous):

ohh see what yore saying i just graphed the line its a tangent itself

OpenStudy (solomonzelman):

tangent line to function \(\large\color{black}{ \displaystyle f(x)=mx+b }\) at \(\large\color{black}{ \displaystyle x=a }\) \(\large\color{black}{ \displaystyle f'(x)=m }\) \(\large\color{black}{ \displaystyle f'(a)=m }\) point on the function when x=a, is also the same point on the tangent line. \(\large\color{black}{ \displaystyle f(a)=ma+b }\) point (a, ma+b) \(\large\color{black}{ \displaystyle y-(ma+b)=m(x-a) }\) \(\large\color{black}{ \displaystyle y=m(x-a)+(ma+b) }\) \(\large\color{black}{ \displaystyle y=mx-ma+ma+b}\) \(\large\color{black}{ \displaystyle y=mx+b}\)

OpenStudy (solomonzelman):

congrants that same tangent line to any linear function at any x=a.

OpenStudy (anonymous):

so the answer would be itself just to confirm

OpenStudy (solomonzelman):

yes.

OpenStudy (anonymous):

thanks for your help

OpenStudy (solomonzelman):

for any curve f(x) if it is linear, and at any x=a.

OpenStudy (solomonzelman):

yw

OpenStudy (anonymous):

what would be the antiderivate of f(x)=-x^2+2

OpenStudy (solomonzelman):

integrate it.

OpenStudy (solomonzelman):

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}x^n~dx=\frac{x^{n+1}}{n+1}}\) (roughly)

OpenStudy (solomonzelman):

and \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}ax^n~dx=\frac{ax^{n+1}}{n+1}}\)

OpenStudy (anonymous):

oh thats a lot I'm asking because they're asking me to graph f(x) but they gave me f'(x) the equation of f '(x) is -x^2+2

OpenStudy (anonymous):

i just figured it would be a tad more simpler than integration

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