I have been stumped on a homework problem for quite some time and I'm hoping to get some help with it.
The line from the origin to the point $(a, f(a))$ on the graph of $f(x) = \frac1{x^2}$ is perpendicular to the tangent line to that graph. What is $a$?
I also need to do the exact same problem for $f(x) = 4/x$.
I don't want the answers without knowing how to actually solve the problem. I know that the derivative of $f(x)$ is equal to $-2/x^3$.
Where do I go from here? We do not have a calculus book here at our university and instead get to use "notes" made by a professor here, which happen to not have any examples at all, nor a way to check the results.
The help is much appreciated! If there's any more info I can give, please let me know.
The condition that the lines be perpendicular is equivalent to the condition that the slope of the tangent be equal to the negative reciprocal of the slope of the radial line from the origin. The slope of the tangent is $f'(a)$ and the slope of the radial line is $f(a)/a$. Thus the general problem comes down to solving the equation $$f'(a) = - a \,\big/\, f(a)\,.$$
Since $f(x)=1/x^2$ is given in the OP's first problem, we have $f(a) = 1/a^2$ and $f'(a) = -2/a^3$. Treating $a$ as an unknown, the equation becomes $${-2 \over a^3} = -{a \over 1/a^2}\,,$$ which has solutions $a = \pm 2^{1/6}$.
For the second problem, setting $f(a) = 4/a$ etc. leads to $a = \pm 2$.