problem+19

Type in the content of your new page here. This is problem 19 from page 295 -- just an example.(Ms. Sweeney) f(x) = x⁵ -5x + 3, so f'(x) = 5x⁴ -5. If we set this equal to zero, we get x = 1 and x = -1 as solutions. Therefore, by the first derivative test there is a local maximum at f(-1) = 7 and there is a local minimum at f(1) = -1.
 * interval || f'(x) || f (x) ||
 * (-∞, -1) || positive || increasing ||
 * (-1, 1) || negative || decreasing ||
 * (1, ∞) || positive || increasing ||

If we were to use the second derivative test, we would find f " (x) = 20x³. f " (-1) = -20; since the second derivative is negative at this point, we know that we have a local maximum at f (-1) = 7. Also, f " (1) = 20; since the second derivative is positive at this x-value, we have a local minimum at f (1) = -1.