Let f be the function given by f(x)=2x^3+3x^2+1 . What is the absolute maximum value of f on the closed interval [−3,1] ?

Answer 1

Hi Jimin,

for this type of problem you need to find the max value within the range for the given function. This might be obvious if you graph the problem to plug in the boundaries and you can see that the boundary is the solution at x=1, where f(1) = 6.

if this is a question on an exam, you might have to show that the two bumps on the function are not the max value.

1. f(x) = 2x³ + 3x² + 1
2. take the derivative f'(x) = 6x² + 6x
3. find the critical values f'(x) = 0 = 6x² + 6x
4. solving for x we get x = 0 and x = -1
5. take the 2nd derivative f"(x) = 12x + 6
6. plug in the critical values in this equation f"(0) = 6 and f"(-1) = -6
7. what we care about is the negative or positive, + U and - ∩ and this shows at these points if these bumps go up or down. note setting the second derivative equal to zero and solving will show the location of the transition point, here it is - 0.5.
8. plug into original equation the critical points f(0) = 1 and f(-1) = 2.
9. these points are not larger than the end points and the max value again is 6 at the x =1 point. this will make sense because you can then draw a quick graph of the plot for an idea of what is happening.

note: This idea can be applied to larger polynomials but is more difficult if the finding the roots is complicated.

I hope this helps, good luck!