Video: Finding the Point of Inflection of a Function Involving Using the Product Rule with Logarithmic Functions If Any

Find (if any) the inflection points of 𝑓(π‘₯) = 3π‘₯Β² ln 2π‘₯.

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Video Transcript

Find, if any, the inflection points of 𝑓 of π‘₯ equals three π‘₯ squared times the natural log of two π‘₯.

To find the points of inflection, we’ll evaluate the second derivative of our function and set it equal to zero. Notice that our function is itself the product of two functions. So we’ll need to use the product rule to differentiate it. This says that for two differentiable functions, 𝑒 and 𝑣, the derivative of their product is 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. We let 𝑒 be equal to three π‘₯ squared and 𝑣 be equal to the natural log of two π‘₯. Then d𝑒 by dπ‘₯ is six π‘₯ and d𝑣 by dπ‘₯ is one over π‘₯. So 𝑓 prime the first derivative of our function is three π‘₯ squared times one over π‘₯ plus six π‘₯ times the natural log of two π‘₯ or three π‘₯ plus six π‘₯ times the natural log of two π‘₯. We’re going to differentiate this again.

We use the product rule to find that the derivative of six π‘₯ times the natural log of two π‘₯ is six plus six times the natural log of two π‘₯. And the second derivative is nine plus six times the natural log of two π‘₯. Let’s set this equal to zero. To solve for π‘₯, we subtract nine and then divide through by six. We raised both sides as the power of 𝑒. And then we divide through by two. So potentially, there’s an inflection point at π‘₯ equals one-half 𝑒 to the negative three over two. But we ought to check whether this is actually an inflection point by checking the values of 𝑓 double prime or the second derivative to either side of this.

A half 𝑒 to the negative three over two is approximately 0.112. So let’s try π‘₯ equals 0.1 and π‘₯ equals 0.12. The second derivative 𝑓 double prime of 0.1 is less than zero. And 𝑓 double prime of 0.12 is greater than zero. The curve goes from being concave down to being concave up. And we can say we do indeed have a point of inflection at π‘₯ equals one-half 𝑒 to the negative three over two. Substituting this value of π‘₯ into original function, we get negative nine over eight 𝑒 cubed. 𝑓 has an inflection point at 𝑒 to the negative three over two over two, negative nine over eight 𝑒 cubed.

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