and at x = − y′ = . This is the larger of the two, so the maximum slope is at x = −. Now we just have to find the y-coordinate, which we get by plugging x into the original
equation: y = . Therefore, the maximum slope is at the point .SOLUTIONS TO PRACTICE PROBLEM SET 11
- Minimum at ( , −6 − 6); Maximum at (− , 6 − 6); Point of inflection at (0, 6).
First, let’s find the y-intercept. We set x = 0 to get y = (0)^3 − 9(0) − 6 = −6. Therefore, the y-intercept is (0, −6). Next, we find any critical points using the first derivative. The derivativeis = 3x^2 − 9. If we set this equal to zero and solve for x, we get x = ± . Plug x = andx = − into the original equation to find the y-coordinates of the critical points: When x = , y = ( )^3 − 9( ) − 6 = −6 − 6. When x = − , y = (− )^3 − 9(− ) − 6 = 6− 6. Thus, we have critical points at ( , −6 − 6) and (− , 6 − 6). Next, we take thesecond derivative to find any points of inflection. The second derivative is: = 6x, whichis equal to zero at x = 0. Note that this is the y-intercept (0, −6), which we already found, sothere is a point of inflection at (0, −6). Next, we need to determine if each critical point ismaximum, minimum, or something else. If we plug x = into the second derivative, the valueis obviously positive, so ( , −6 − 6) is a minimum. If we plug x = − into the secondderivative, the value is obviously negative, so (− , 6 − 6) is a maximum. Now, we candraw the curve. It looks like the following: