Video Transcript
The figure shows the graph of 𝑓 of
𝑥 equals negative 𝐴 over 𝑥 squared plus 𝐵 over 𝑥 cubed for positive constants
𝐴, 𝐵. Find the exact values of the
constants if the local minimum value shown is negative one and the inflection point
𝑃 occurs when 𝑥 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 𝑥
and 𝑓 double prime of 𝑥 then set them equal to zero, we’ll be able to solve for 𝐴
and 𝐵. Before we do, though, let’s rewrite
our function for 𝑓 of 𝑥. One over 𝑥 squared is 𝑥 to the
power of negative two. And one over 𝑥 cubed is 𝑥 to the
power of negative three. So 𝑓 of 𝑥 is negative 𝐴𝑥 to the
power of negative two plus 𝐵𝑥 to the power of negative three.
We can then use the power rules for
differentiation to find an expression for 𝑓 prime of 𝑥. Suppose we have some general
expression 𝑎𝑥 to the 𝑛th power. The derivative of this expression
with respect to 𝑥 is then 𝑛 times 𝑎 times 𝑥 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
𝐴𝑥 to the power of negative two with respect to 𝑥, we get negative two times
negative 𝐴𝑥 to the power of negative three.
And differentiating the second part
of our function, we get negative three times 𝐵𝑥 to the power of negative four. And then we can simplify. And we find 𝑓 prime of 𝑥 is two
𝐴𝑥 to the power of negative three minus three 𝐵𝑥 to the power of negative
four. Differentiating each term once more
then gives us an expression for 𝑓 double prime of 𝑥. It’s negative six 𝐴𝑥 to the power
of negative four plus 12𝐵𝑥 to the power of negative five.
We also know that since an
inflection point 𝑃 occurs when 𝑥 is equal to two, when 𝑥 is two 𝑓 double prime
of 𝑥 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 12𝐵 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 𝐵. If we divide through by
three-eighths, then this equation can be written as zero equals negative 𝐴 plus
𝐵. This is equivalent to saying 𝐴 is
equal to 𝐵.
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 𝐴𝑥 to the power of negative
three minus three 𝐵𝑥 to the power of negative four is equal to zero. Since we have established that 𝐴
is equal to 𝐵, let’s rewrite this as zero equals two 𝐴 times 𝑥 to the power of
negative three minus three 𝐴 times 𝑥 to the power of negative four.
Then since 𝐴 is a constant, we can
factor this expression to solve for 𝑥. We’re going to take a factor of
𝐴𝑥 to the power of negative four outside of our parentheses. Then on the inside of the
parentheses, we have two 𝑥 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 𝐴𝑥 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 𝐵 are both positive, so 𝐴 cannot be equal to zero.
Instead, we’re going to solve the
equation two 𝑥 minus three equals zero to find the 𝑥-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 𝑥 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 𝑥 equals three
over two, is in fact negative one. So we’re going to substitute 𝑥
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 𝐵 times
three over two to the power of negative three. Once again, let’s replace 𝐵 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 𝐵 and 𝐴 were equal
to one another. So we can also say that 𝐵 is 27
over four. So 𝐴 is 27 over four, and 𝐵 is 27
over four.
