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 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,
$\displaystyle \frac{y-2}{x-1}=\frac{1}{x^{3}}$
You could simplify the expression as you wish(eg; take y=f(x) if you want) to get the equation of the curve.
Hope this will help you.
Hi cole123
$\displaystyle \frac{d}{dx}\left(-\frac{1}{2x^2}+C\right)=\frac{d}{dx}\left(-\frac{x^{-2}}{2}+C\right)=x^{-3}$
If f(1)=2
$\displaystyle -\frac{1}{2}+C=2\ \Rightarrow\ C=2.5$
$\displaystyle f(x)=2.5-\frac{1}{2x^2}$
If we integrate $\displaystyle \frac{1}{x^3}$
we get $\displaystyle \frac{x^{-2}}{-2}+C$
Hi cole123,
It's because, when we integrate, we must calculate the new power
before dividing by it.
$\displaystyle \int{x^n}dx=\frac{x^{n+1}}{n+1}+C$
and when we differentiate, we multiply by the original power
before reducing the power by 1.
$\displaystyle \frac{d}{dx}\left(x^n+C\right)=nx^{n-1}$
Hi Sudharaka,
yes, with the curve, only the point of tangency is on the line
in that vicinity.