Video: Finding the Concavity of a Curve Given the Derivative

Let 𝑓 be a function whose derivative is given by 𝑓′(π‘₯) = 5π‘₯/(2π‘₯Β² + 6)Β². On which of the following intervals of π‘₯ is the graph of 𝑓 concave up? [A] (0, ∞) [B] (βˆ’βˆž, 0) [C] (βˆ’βˆž, βˆ’1) βˆͺ (1, ∞) [D] (βˆ’1, 1 )

04:15

Video Transcript

Let 𝑓 be a function whose derivative is given by 𝑓 prime of π‘₯ equals five π‘₯ over two π‘₯ squared plus six all squared. On which of the following intervals of π‘₯ is the graph of 𝑓 concave up? Is it a) the open interval from zero to ∞. B) The open interval negative ∞ to zero. Is it c) the union of the open interval from negative ∞ to negative one and the open interval one to ∞? Or d) the open interval negative one to one.

We’re going to need to be a little bit careful here. We have not been given an expression for the function 𝑓 itself, but one for its derivative, 𝑓 prime of π‘₯. We’re looking to find the intervals on which the graph of 𝑓 is concave up. So what does this mean?

When a graph is concave up, we say its first derivative, 𝑓 prime of π‘₯, is increasing. We can consider the rate of change of that first derivative. That’s the derivative of the derivative, 𝑓 double prime of π‘₯. And what this means is 𝑓 double prime of π‘₯ must be greater than zero. So to answer this question, we’re going to need to differentiate 𝑓 prime of π‘₯ to find the second derivative and then establish on which intervals of π‘₯ that is greater than zero.

Now 𝑓 prime of π‘₯ is the quotient of two functions. So we use the quotient rule. And this says that the derivative of the quotient of two differentiable functions, 𝑒 and 𝑣, is 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared. The numerator of our fraction is five π‘₯. So we’re going to let 𝑒 be equal to five π‘₯. And 𝑣 is the denominator. It’s two π‘₯ squared plus six squared.

It should be quite clear from our formula for the quotient rule that we’re going to need to differentiate each of these functions with respect to π‘₯. Well, the derivative of five π‘₯ is fairly straightforward. It’s simply five. But what about the derivative of this composite function, two π‘₯ squared plus six all squared?

Now we could use the chain rule. Or we can use a special case of the chain rule, which is called the general power rule. What we do is we multiply the entire inner function by the value of the exponent, so by two. And then we reduce that exponent by one. Then we multiply that by the derivative of the inner function. Well, the derivative of two π‘₯ squared plus six with respect to π‘₯ is four π‘₯. So we find that d𝑣 by dπ‘₯ in its simplest form is eight π‘₯ times two π‘₯ squared plus six.

Let’s substitute everything we have into the quotient rule. The derivative of 𝑓 prime of π‘₯, which is 𝑓 double prime of π‘₯, is 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared. We can simplify the denominator of this expression by multiplying the exponent. So we get two π‘₯ squared plus six to the fourth power. On our numerator, we can factor out two π‘₯ squared plus six. And so we find that 𝑓 double prime of π‘₯ is two π‘₯ squared plus six times five times two π‘₯ squared plus six minus 40π‘₯ squared over two π‘₯ squared plus six to the fourth power.

We distribute the first set of parentheses to give us 10π‘₯ squared plus 30. And so we see this simplifies to two π‘₯ squared plus six times 30 minus 30π‘₯ squared all over two π‘₯ squared plus six to the fourth power. Now we say that, for the graph to be concave up, the second derivative must be greater than zero. So we need to find the intervals of π‘₯ for which this is the case. And this might look quite complicated to start.

However, we know that any real number to the fourth power is greater than zero. We know two π‘₯ squared is greater than zero. And therefore, two π‘₯ squared plus six is greater than zero. So we need to work out where the other part of the expression 30 minus 30π‘₯ squared is greater than zero. Then we have the product of two positives divided by a positive, which we know to be a positive.

Let’s clear some space and solve this inequality. We’ll divide through by 30. And that gives us one minus π‘₯ squared is greater than zero. Let’s factor the expression one minus π‘₯ squared and for now just set it equal to zero. We get one minus π‘₯ times one plus π‘₯ equals zero. And for this to be true, either one minus π‘₯ must be equal to zero or one plus π‘₯ must be equal to zero. And so we see that when π‘₯ is equal to one and negative one, the expression one minus π‘₯ squared is equal to zero. So where is it greater than zero?

Well, there are a number of ways we can achieve this. But one way is to draw the graph. Now we see the graph of 𝑦 equals one minus π‘₯ squared is greater than zero here. That’s values of π‘₯ greater than negative one and less than one. This means our second derivative is greater than zero for values of π‘₯ on the open interval negative one to one. And in turn, the graph of 𝑓 is concave up over that same interval. So the correct answer is d) the open interval negative one to one.

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