Video: Finding Intervals of Upward and Downward Concavity of a Function Involving Roots

Determine the intervals on which the function 𝑓(π‘₯) = βˆ’3π‘₯ + √(9π‘₯Β² + 1) is concave up and down.

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Video Transcript

Determine the intervals on which the function 𝑓 of π‘₯ equals negative three π‘₯ plus the square root of nine π‘₯ squared plus one is concave up and down.

Let’s begin by recalling what it means for a function to be either concave up or down. When a function is concave up, the tangents to the graph of the function lie below the graph itself. We see that the slope of these tangents is increasing. In the sketch, the slope is changing from negative to zero to positive. And so, the value of the slope is getting larger. We can say then that when a function is concave up, its first derivative 𝑓 prime of π‘₯ is increasing. If 𝑓 prime of π‘₯ is increasing, then its own derivative 𝑓 double prime of π‘₯ must be positive. So, we can also conclude that when a function is concave up, its second derivative is greater than zero.

When a function is concave down however, the opposite is true. The tangents to the graph lie above the graph itself. The first derivative 𝑓 prime of π‘₯ is decreasing. In the sketch, we see it’s changing from positive to zero to negative. And so, the second derivative 𝑓 double prime of π‘₯ will be negative. To answer this question then, we’re going to need to find an expression for the second derivative of our function 𝑓 of π‘₯. So, we’re going to need to differentiate twice.

Before we do though, we may find it easier to write the second part of the definition of 𝑓 of π‘₯, that’s the square root of nine π‘₯ squared plus one, as nine π‘₯ squared plus one to the power of one-half. Now, we can differentiate to find an expression for the first derivative 𝑓 prime of π‘₯. Using the power rule of differentiation, the derivative of negative three π‘₯ is negative three. But what about the derivative of nine π‘₯ squared plus one to the power of one-half?

Well, this is where we need to use the chain rule extension to the power rule, which tells us that if we have some function 𝑔 of π‘₯ to the power of 𝑛, and we want to find its derivative with respect to π‘₯. Then, this is equal to 𝑛 multiplied by 𝑔 prime of π‘₯ multiplied by 𝑔 of π‘₯ to the power of 𝑛 minus one. We multiply by the exponent and reduce the exponent by one. But we also multiply by the derivative of our function 𝑔 of π‘₯.

So, differentiating nine π‘₯ squared plus one to the power of one-half then. We have 𝑛, that’s the exponent, which is one-half. Then, the derivative of nine π‘₯ squared plus one, which is 18π‘₯. And then nine π‘₯ square plus one to the power of 𝑛 minus one, which will be negative one-half. This all simplifies to negative three plus nine π‘₯ multiplied by nine π‘₯ squared plus one to the power of negative one-half. So, we have our expression for the first derivative 𝑓 prime of π‘₯. But now, we need to differentiate again.

The derivative of the first term negative three is straightforward. It’s just zero as negative three is a constant. But what about the derivative of nine π‘₯ multiplied by nine π‘₯ square plus one to the power of negative one-half. Well, here, we have a product of differentiable functions. So, we’re going to need to apply the product rule. This tells us that for two differentiable functions 𝑒 and 𝑣, the derivative with respect to π‘₯ of their product 𝑒𝑣 is equal to 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. We multiply each function by the derivative of the other.

So, we can define 𝑒 to be the first factor, that’s nine π‘₯, and 𝑣 to be the second factor, that’s nine π‘₯ squared plus one to the power of negative one-half. Using the power rule of differentiation, d𝑒 by dπ‘₯ is equal to nine. And then, to find d𝑣 by dπ‘₯, we need to use the chain rule extension to the power rule again. Which tells us that d𝑣 by dπ‘₯ will be equal to negative a half multiplied by 18π‘₯ multiplied by nine π‘₯ squared plus one to the power of negative three over two. We’ve reduced the exponent by one.

Using the product rule then, we have that the derivative with respect to π‘₯ of 𝑒𝑣 is equal to 𝑒 times d𝑣 by dπ‘₯. That’s nine π‘₯ multiplied by negative a half 18π‘₯ multiplied by nine π‘₯ squared plus one to the negative three over two. Plus 𝑣 times d𝑒 by dπ‘₯, that’s nine multiplied by nine π‘₯ squared plus one to the power of negative one-half. We can then take out a shared factor of nine multiplied by nine π‘₯ squared plus one to the power of negative three over two.

What remains in the first term is negative nine π‘₯ squared. And what remains in the second is nine π‘₯ squared plus one to the power of one. Because, here, when we add the exponents of one and negative three over two, we get negative one-half. The largest set of parentheses simplifies nicely. We have negative nine π‘₯ squared plus nine π‘₯ square plus one, which all just simplifies to one. So, our derivative is simply nine multiplied by nine π‘₯ squared plus one to the power of negative three over two. And applying laws of exponents, we can rewrite this as nine over the square root of nine π‘₯ squared plus one cubed.

So, returning to where we were writing down our expression for 𝑓 double prime of π‘₯ then, we have that the second derivative is equal to zero plus nine over the square root of nine π‘₯ squared plus one cubed. And of course, we don’t need to include the zero.

Remember that the purpose of this question is to determine where the function 𝑓 of π‘₯ is concave up and concave down. And we recall that we said a function’s concave up when its second derivative is greater than zero. So, let’s first consider where this is true. Well, this will be the case when nine over the square root of nine π‘₯ squared plus one cubed is greater than zero. Let’s consider this quotient. The numerator of this quotient is nine, which is a constant greater than zero. In the denominator, nine π‘₯ square plus one will give a value greater than or equal to one. And when we take its square root, by definition, we’re taking the positive square root, so we have a positive value. We’re then cubing this, which will give another positive value.

So, we see that, for all values of π‘₯, this quotient will be a positive value divided by a positive value, which will return a positive value. This means that the second derivative 𝑓 double prime of π‘₯ is, in fact, always positive. And therefore, the function 𝑓 of π‘₯ will be concave up throughout its domain, which in this case is the entire set of real numbers. Using interval notation then, we can conclude that the function 𝑓 of π‘₯ will be concave up on the open interval from negative ∞ to ∞, and there is no interval on which the function 𝑓 of π‘₯ is concave down.

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