Determine the local maximum and local minimum values of 𝑓 of 𝑥 equals four 𝑥 cubed minus 12𝑥 minus five.
Local maxima and local minima are examples of critical points. And we know that the critical points of a function occur when its first derivative is equal to zero or is undefined. We, therefore, need to find an expression for the first derivative of our function 𝑓 prime of 𝑥 and then find where it’s equal to zero. In order to find 𝑓 prime of 𝑥, we can use the power rule of differentiation. 𝑓 prime of 𝑥 is equal to four multiplied by three 𝑥 squared minus 12 multiplied by one. Remember, the derivative of a constant, in this case negative five, is just zero, which simplifies to 12𝑥 squared minus 12. So we have our expression for the first derivative of our function 𝑓 of 𝑥.
Next, we set this expression equal to zero and solve the resulting equation for 𝑥. Adding 12 to both sides and then dividing by 12 gives 𝑥 squared is equal to one. We solve by square routing, remembering that we must take both the positive and the negative values. 𝑥 is equal to plus our minus the square root of one. And as the square root of one is just one, we have that 𝑥 is equal to plus or minus one. Now, these are the 𝑥-values at which the critical points of our function occur. We also need to determine the values of the function itself at these points. To do so, we substitute each 𝑥-value in turn into our function 𝑓 of 𝑥.
𝑓 of one is equal to four multiplied by one cubed minus 12 multiplied by one minus five, which is equal to negative 13. 𝑓 of negative one is four multiplied by negative one cubed minus 12 multiplied by negative one minus five, which is equal to three. So we find that our function has critical points at the points one, negative 13 and negative one, three. But the question asks us to determine the local maximum and local minimum values of our function 𝑓 of 𝑥. So we need to classify these critical points.
In order to do this, we need to apply the second derivative test. We’ll find an expression for the second derivative 𝑓 double prime of 𝑥 of our function and then evaluate this at each of the critical points. Depending on the sign of 𝑓 double prime of 𝑥 at each critical point, this will tell us whether there are local maximum, a local minimum, or perhaps a point of inflection. Differentiating our expression 𝑓 prime of 𝑥 again, remember, 𝑓 prime of 𝑥 was equal to 12𝑥 squared minus 12. We see that the second derivative 𝑓 double prime of 𝑥 is equal to 12 multiplied by two 𝑥, which is 24𝑥.
Evaluating this when 𝑥 is equal to one gives 24 multiplied by one, which is 24. This is greater than zero. And the second derivative test tells us that if the second derivative of a function is positive at a critical point, then that critical point is a local minimum. So we know that our critical point one, negative 13 is a local minimum of the function 𝑓 of 𝑥. Evaluating 𝑓 double prime of 𝑥 at our other critical point when 𝑥 is equal to negative one, we obtained negative 24. This is less than zero and so the second derivative test tells us that this critical point is a local maximum of the function 𝑓 of 𝑥.
We can conclude them that this function 𝑓 of 𝑥 has a local maximum value of three at 𝑥 equals negative one and a local minimum value of negative 13 at 𝑥 equals positive one. Remember, in this question, we use differentiation to find the first derivative of our function, we will call that critical points. The first derivative will be equal to zero or be undefined. We evaluated the function itself at each of our critical points. And then we used the second derivative test to classify these critical points as local minima or local maxima.