 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# The Second Derivative Test for Relative Maximum and Minimum - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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It is also possible to find relative maxima and minima by using the second derivative of a function. First, calculate the first derivative of the function and find its critical points. Then, calculate the second derivative of the function. Plug in the critical points found into the second derivative. If the result is positive, the graph is concave up at that point, so it is a relative minimum. If the result is negative, the graph is concave down at that point, so it is a relative maximum.

If the second derivative is 0 at a critical point, then the second derivative test has failed, and you must use the first derivative test to determine if that point is a maximum or minimum.

We're finding relative maxima, and minima using the second derivative test. Here is a problem. Find the relative relative maxima, and minima of the function h(x) equals 9x times e to the -x over 3.

Now the first step is to find the first and second derivatives of this function. So I'll need the product rule to differentiate this. H'(x) will be the first function 9x, times the derivative of the second. I'll need the chain rule for that.

The derivative of e to the -x over 3, is -1/3 e to the -x over 3. It's e to the -x over 3 times the derivative of -x over 3. That's where the -1/3 comes from. Plus the second function; e to the -x over 3 times the derivative of the first, and that's just 9.

Now let me factor this. It will be easier to differentiate a second time if I do that. There's an e to the -x over 3 in both of these terms. So I'll factor that out. What will I be left with? 9x times -1/3 is -3x. Then here I just have a +9. So I have another product to two functions. I have to differentiate that to get h''.

So it's going to be first times the derivative of the second. So that's -3x plus 9, times the derivative of the second. This is exactly the same derivative I just took, this one. So -1/3e to the -x over 3, plus the second; e to the -x over 3, times the derivative of the first. The derivative of this linear function is -3.

So once again, e to the -x over 3, is a common factor to both terms. It's in the first term, it's in the second term. So I factor that out, what's left? Well I have a -3x times -1/3; that's x. I have a 9 times -1/3, that's -3. I have another -3 over here. So I have x minus 6 times e to the -x over 3. That's my second derivative.

Second step; find critical points. This is going to involve the first derivative. I want to find out whether the first derivative is 0, or undefined. Now again the first derivative is -3x plus 9 times e to the -x over 3. This derivative is never going to be undefined. Neither of these factors will ever be undefined. So we need to focus on where the derivative could equal 0. It will equal 0 only when this factor equals 0. This thing is never going to equal to 0. E to any power always positive.

So -3x plus 9 equals 0 when x equals 3. That's our critical point. Then the last step is to compute the second derivative at each critical point. There's only one, so I need to plug 3 into the second derivative that I found down here. This is my second derivative; (x minus 6)e to the -x over 3. So I'm plugging 3 in, and I get (3-6) e to the minus 3 over 3 which is -3 times e to the -1.

Now again, e to any power, even a negative power, is always going to be a positive number. So this is positive times the negative number is negative. All I really need to know is that this quantity is negative, because the second derivative says, if the first derivative is 0 at a point, and the second derivative is negative, let's draw a little picture. It's concave down. It's got a horizontal tangent. We have a relative maximum. So h has a relative maximum at x equals 3. That's the second derivative test.