Given a function $g(x)$, and a function $f(x)$ are real, continuous functions on all reals is a there a method to prove whether a function is always above lets say my toolkit is first, and second derivatives, and I would like to avoid graphing. I would always consider expanding my toolkit to solve this problem is just that I don't want to rely on graphing always to have the say of the solution to the problem which function is above or below. Thus, said is there a way to analytically prove a function is above another function for a given interval? The given interval of course is a subset of all Reals.
Right now what I do solve this problem is the following I graph the end points of a problem and lets say I need to find area, or volume knowing which one is top, and bottom I get the intersection points and find a point in the middle and see which one is on top, and which one is on the bottom.
Proving a function is always above another function
calculusfunctions
Best Answer
Look at $h(x)=f(x)-g(x)$. Take the derivative $$h'(x)=f'(x)-g'(x)$$and set it to zero, to find out where there is a maximum/minimum. You also need to consider the values of $h(x)$ at the ends of the interval $[a,b]$. So you have several cases. What you are looking for is $h(a)>0$, $h(b)>0$ first. Then see if there is a minimum of $h(x)$ in your interval. If there is, calculate $h(x)$ at all roots $x_i$ of $h'(x)=0$ in your interval. If all $h(x_i)$ are positive, then $f(x)>g(x)$ in the given interval.