Calculus Problem of the Week July 18, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week.

This week’s problem:
(click “see the solution” at the bottom of post to, well, see the solution.)
The function f is a continuous function whose second derivative is a non-zero constant and the equation of the tangent line to f at x=2 is y=12x-12. If the function f itself is not a constant and has only one term, find f.
See the solution.

There is a lot of information about f given in the problem, so a good first step is to try to pull even more information out. Lets look at each statement:

  • The function f is a continuous function whose second derivative is a non-zero constant

    • This means that if we take the derivative twice we end up with a number like 5 or -2. What functions give us a constant other than zero when we take their derivative? LINEAR FUNCTIONS. So, the first derivative must be linear! Ok we are making progress. Now, what type of function gives you a linear function when you take the derivative? A QUADRATIC like ax^2+bx+c.

  • The equation of the tangent line to f at x=2 is y=12x-12

    • This means that f'(2)=12

  • The function f itself is not a constant and has only one term

    • We already figured out that f would look something like ax^2+bx+c since it must be quadratic. But, if it only one term that means that f=ax^2 for some number a that is not zero. We already know that f'(2)=12 so I will find the derivative of our f so far: f'=(ax^2)'=2ax. Finally since this must be 12 when x=2, 4a=12 and therefore a=3 giving us f=3x^2