# Thread: [SOLVED] Finding the equation of a function

1. ## [SOLVED] Finding the equation of a function

find the function f(x) the graph passes through (1,2) and whose tangent line at the point (x,F(x)) has a slope of 1/x3

2. Originally Posted by cole123
find the function f(x) the graph passes through (1,2) and whose tangent line at the point (x,F(x)) has a slope of 1/x3
Dear cole,

What is the slope of the graph? Is it $\frac{1}{3} ?$ There is a typo.

3. sorry it is 1/x^3

4. Dear cole123,

Suppose (x,y) is any point on this curve. Then, since (1,2) is also on this curve, the tangent of the curve is,

$\frac{y-2}{x-1}=\frac{1}{x^{3}}$

You could simplify the expression as you wish(eg; take y=f(x) if you want) to get the equation of the curve.

5. Im having trouble with that part I know the next step leads to:

f(x)= -(1/2)x^-2 + c but I have no idea how to get this!

6. Hi cole123

$\frac{d}{dx}\left(-\frac{1}{2x^2}+C\right)=\frac{d}{dx}\left(-\frac{x^{-2}}{2}+C\right)=x^{-3}$

If f(1)=2

$-\frac{1}{2}+C=2\ \Rightarrow\ C=2.5$

$f(x)=2.5-\frac{1}{2x^2}$

If we integrate $\frac{1}{x^3}$

we get $\frac{x^{-2}}{-2}+C$

7. I dont understand why it is not -(1/3) x^-2?

8. Originally Posted by cole123
Im having trouble with that part I know the next step leads to:

f(x)= -(1/2)x^-2 + c but I have no idea how to get this!
Dear cole123,

My method is incorrect. I had done it assuming the curve is a straight line which is of course incorrect. But Archie Meade's answer is correct.

Thank you for giving the correct answer. Now I know what I did wrong.

9. Hi cole123,

It's because, when we integrate, we must calculate the new power
before dividing by it.

$\int{x^n}dx=\frac{x^{n+1}}{n+1}+C$

and when we differentiate, we multiply by the original power
before reducing the power by 1.

$\frac{d}{dx}\left(x^n+C\right)=nx^{n-1}$

Hi Sudharaka,
yes, with the curve, only the point of tangency is on the line
in that vicinity.

10. Thank you very much!