get y = −x^4 + 13x^2 − 36. Now we can take the derivative: = −4x^3 + 26x. Next, we set thederivative equal to zero to find the critical points. There are three solutions: x = 0, x = ,and x = −.We plug these values into the original equation to find the y-coordinates of the critical points:
When x = 0, y = −(0)^4 + 13(0)^2 − 36 = −36. When x = , y = + 13 −36 = .
When, x = − , y = + − 36 = .
Thus, we have critical points at (0, −36), , and .
Next, we take the second derivative to find any points of inflection. The second derivative is
= −12x^2 + 26, which is equal to zero at x = and x = − . We plug these valuesinto the original equation to find the y-coordinates.When x = , then y = − − 36 = −.
When x = − , then y = − − 36 = − . So there are points of
inflection at and . Next, we need to determine if eachcritical point is maximum, minimum, or something else. If we plug x = 0 into the secondderivative, the value is positive, so (0, −36) is a minimum. If we plug x = − into the