Question Video: Identifying the Equation of a Graph Given Its Local Minimum and Point of Inflection Mathematics • Higher Education

The figure shows the graph of 𝑓(π‘₯) = (βˆ’π΄/π‘₯Β²) + (𝐡/π‘₯Β³) for positive constants 𝐴, 𝐡. Find the exact values of the constants if the local minimum value shown is βˆ’1 and the inflection point 𝑃 occurs when π‘₯ = 2.

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Video Transcript

The figure shows the graph of 𝑓 of π‘₯ equals negative 𝐴 over π‘₯ squared plus 𝐡 over π‘₯ cubed for positive constants 𝐴, 𝐡. Find the exact values of the constants if the local minimum value shown is negative one and the inflection point 𝑃 occurs when π‘₯ equals two.

In order to answer this question, we need to recall what it means for the graph of a function to have a local minimum or an inflection point.

Let’s begin by thinking about relative minima and maxima. Suppose we have some continuous function. A minimum or a maximum will occur at places where the derivative 𝑓 prime is equal to zero or does not exist. It can also occur at the endpoints of the function, but we’re not interested in that in this case. Then a point of inflection is a point where the concavity of the function changes. At these points, either the second derivative is equal to zero or it’s undefined.

And so this is really useful because if we can take our original function and find expressions for 𝑓 prime of π‘₯ and 𝑓 double prime of π‘₯ then set them equal to zero, we’ll be able to solve for 𝐴 and 𝐡. Before we do, though, let’s rewrite our function for 𝑓 of π‘₯. One over π‘₯ squared is π‘₯ to the power of negative two. And one over π‘₯ cubed is π‘₯ to the power of negative three. So 𝑓 of π‘₯ is negative 𝐴π‘₯ to the power of negative two plus 𝐡π‘₯ to the power of negative three.

We can then use the power rules for differentiation to find an expression for 𝑓 prime of π‘₯. Suppose we have some general expression π‘Žπ‘₯ to the 𝑛th power. The derivative of this expression with respect to π‘₯ is then 𝑛 times π‘Ž times π‘₯ to the power of 𝑛 minus one. We multiply by the power and then reduce it by one. And when we’re working with a function that has more than one term, we can just do this term by term. So if we differentiate negative 𝐴π‘₯ to the power of negative two with respect to π‘₯, we get negative two times negative 𝐴π‘₯ to the power of negative three.

And differentiating the second part of our function, we get negative three times 𝐡π‘₯ to the power of negative four. And then we can simplify. And we find 𝑓 prime of π‘₯ is two 𝐴π‘₯ to the power of negative three minus three 𝐡π‘₯ to the power of negative four. Differentiating each term once more then gives us an expression for 𝑓 double prime of π‘₯. It’s negative six 𝐴π‘₯ to the power of negative four plus 12𝐡π‘₯ to the power of negative five.

We also know that since an inflection point 𝑃 occurs when π‘₯ is equal to two, when π‘₯ is two 𝑓 double prime of π‘₯ is equal to zero. So we evaluate 𝑓 double prime of two and set that equal to zero. So zero is negative six 𝐴 times two to the power of negative four plus 12𝐡 times two to the power of negative five. Two to the power of negative four is one over 16. And two to the power of negative five is one over 32. So this simplifies to zero equals negative three-eighths 𝐴 plus three-eighths 𝐡. If we divide through by three-eighths, then this equation can be written as zero equals negative 𝐴 plus 𝐡. This is equivalent to saying 𝐴 is equal to 𝐡.

So let’s clear some space and move on to the information about the local minimum. We know this occurs when the derivative is equal to zero, in other words, when two 𝐴π‘₯ to the power of negative three minus three 𝐡π‘₯ to the power of negative four is equal to zero. Since we have established that 𝐴 is equal to 𝐡, let’s rewrite this as zero equals two 𝐴 times π‘₯ to the power of negative three minus three 𝐴 times π‘₯ to the power of negative four.

Then since 𝐴 is a constant, we can factor this expression to solve for π‘₯. We’re going to take a factor of 𝐴π‘₯ to the power of negative four outside of our parentheses. Then on the inside of the parentheses, we have two π‘₯ minus three. For the product of these two expressions to be equal to zero, we know either one or other must be equal to zero. But there is no way for the expression 𝐴π‘₯ to the power of negative four to be equal to zero unless 𝐴 itself is equal to zero. But of course, we’re told that 𝐴 and 𝐡 are both positive, so 𝐴 cannot be equal to zero.

Instead, we’re going to solve the equation two π‘₯ minus three equals zero to find the π‘₯-value at which the local minimum occurs. To do so, we add three to both sides and then divide through by two. So the local minimum occurs at a point where π‘₯ equals three over two. And in fact, we’re almost done. We haven’t yet used the information that the value of the local minimum, the value of the function at π‘₯ equals three over two, is in fact negative one. So we’re going to substitute π‘₯ equals three over two into our original function and set that equal to negative one.

When we do, we get negative one equals negative 𝐴 times three over two to the power of negative two plus 𝐡 times three over two to the power of negative three. Once again, let’s replace 𝐡 with 𝐴 and then solve this equation for 𝐴. The simplest way to do this is to factor 𝐴 from the right-hand side. Then we can evaluate negative three over two to the power of negative two plus three over two to the power of negative three. It’s negative four over 27. So to solve for 𝐴, we’ll divide both sides of this equation by negative four over 27.

Well, negative one divided by negative four over 27 is 27 over four. So we found the value of 𝐴. But remember 𝐡 and 𝐴 were equal to one another. So we can also say that 𝐡 is 27 over four. So 𝐴 is 27 over four, and 𝐡 is 27 over four.

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