# Video: APCALC05AB-P2B-Q06-586158649289

The functions 𝑓 and 𝑔 have continuous second derivatives. The table gives values of the functions and their derivatives as selected values of 𝑥. i. Let ℎ(𝑥) = 𝑓(𝑔(𝑥)). Write an equation for the line tangent to the graph of ℎ at 𝑥 = 4. ii. Let 𝑅(𝑥) = 𝑔(𝑥)/𝑓(𝑥). Find 𝑅′(7). iii. Evaluate ∫^2_1 𝑓″(3𝑥) d𝑥.

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### Video Transcript

The functions 𝑓 and 𝑔 have continuous second derivatives. The table gives values of the functions and their derivatives as selected values of 𝑥. i) Let ℎ of 𝑥 equal 𝑓 of 𝑔 of 𝑥. Write an equation for the line tangent to the graph of ℎ at 𝑥 equals four. ii) Let 𝑅 of 𝑥 equal 𝑔 of 𝑥 over 𝑓 of 𝑥. Find 𝑅 prime of seven. iii) Evaluate the integral from one to two of 𝑓 double prime of three 𝑥 with respect to 𝑥.

Beginning with part i then, we’re asked to write an equation for the line tangent to the graph of ℎ at the point 𝑥 equals four. We first recall that the general equation of any straight line is 𝑦 equals 𝑚𝑥 plus 𝑏, where 𝑚 is the slope of the line and 𝑏 is its 𝑦-intercept. The slope of a tangent is the same as the slope of the curve at that point, which is given by its first derivative. So, in order to find the value of 𝑚, we need to find ℎ prime, that’s the first derivative of ℎ, and then evaluate it when 𝑥 is equal to four.

But we see that ℎ of 𝑥 is a function of a function; it’s 𝑓 of 𝑔 of 𝑥. So, we need to recall how we can find the derivative of a composite function. Well, this is exactly the setting we need in order to apply the chain rule. This tells us that if ℎ of 𝑥 is equal to the composite function 𝑓 of 𝑔 of 𝑥, then its first derivative ℎ prime of 𝑥 is equal to 𝑔 prime of 𝑥 multiplied by 𝑓 prime of 𝑔 of 𝑥. That’s the derivative of the inner function multiplied by the derivative of the outer function with the inner function still inside.

We don’t need a general expression for ℎ prime of 𝑥. We just need to evaluate it when 𝑥 is equal to four. So, we see that ℎ prime of four is equal to 𝑔 prime of four multiplied by 𝑓 prime of 𝑔 of four. And now, we can use the table that we’ve been given. Looking carefully at the table, we first see that 𝑔 prime of four is equal to five. Looking in the 𝑔 of 𝑥 column of our table, we see that 𝑔 of four is equal to seven. So, substituting these two values, we see that ℎ prime of four is equal to five multiplied by 𝑓 prime of seven.

Finally, for this stage we see that 𝑓 prime of seven is equal to three. ℎ prime of four is, therefore, equal to five multiplied by three, which is equal to 15. So, we found the slope of our tangent ℎ prime of four, which is equal to 15. In order to find the full equation of our tangent, we need to know the coordinates of one point that lies on this line. We know that 𝑥 is equal to four, but what is ℎ of 𝑥 at this point? Well, ℎ of four will be equal to 𝑓 of 𝑔 of four. And we now need to use our table again. 𝑔 of four, we already know to be seven. So, ℎ of four is equal to 𝑓 of seven.

From the first column in our table, we see that 𝑓 of seven is equal to negative five. And therefore, the point four, negative five lies on this tangent line. Finally, we substitute this pair of 𝑥-, 𝑦-values into the equation of our tangent. We have 𝑦, negative five, is equal to 𝑚𝑥, that’s 15 times four, plus 𝑏. We can solve this equation by subtracting 15 times four, that’s 60, from each side to give 𝑏 is equal to negative 65. Finally, substituting the value of our slope 𝑚 and the value of 𝑏 into the equation of our tangent, we have that 𝑦 is equal to 15𝑥 minus 65. There are, of course, equivalent forms that we could write the equation of this tangent in.

Now, let’s consider part ii of the question, in which we’re given the function 𝑅 of 𝑥 equals 𝑔 of 𝑥 over 𝑓 of 𝑥 and asked to find 𝑅 prime of seven. That’s the first derivative of 𝑅 evaluated when 𝑥 is equal to seven. We’re, therefore, going to need to find an expression for 𝑅 prime of 𝑥 in terms of 𝑓 of 𝑥, 𝑔 of 𝑥, and their derivatives. And to do so, we note that 𝑅 of 𝑥 is a quotient. It’s the quotient of the two differentiable functions, 𝑔 of 𝑥 and 𝑓 of 𝑥.

We can, therefore, apply the quotient rule, which tells us that for two differentiable functions 𝑔 and 𝑓, the derivative with respect to 𝑥 of 𝑔 over 𝑓 is equal to 𝑓 multiplied by the derivative of 𝑔 minus 𝑔 multiplied by the derivative of 𝑓 over 𝑓 squared. I’ve changed the letters from the way you may usually see them in the quotient rule so that they’re consistent with how 𝑅 of 𝑥 has been defined. So, we now have an expression for the derivative 𝑅 prime of 𝑥. But we actually want to evaluate this at a point.

We have then that 𝑅 prime of seven is equal to 𝑓 of seven multiplied by 𝑔 prime of seven minus 𝑔 of seven multiplied by 𝑓 prime of seven over 𝑓 of seven squared. And we can find all of these values from the table. 𝑓 of seven, first of all, is equal to negative five. 𝑔 prime of seven is equal to nine. 𝑔 of seven is equal to 10. And 𝑓 prime of seven is equal to three. So, we have all the values we need in order to evaluate this derivative. Simplifying gives negative 45 minus 30 over 25. That’s negative 75 over 25, which is exactly equal to negative three. So, by applying the quotient rule and then substituting 𝑥 equals seven, we found that 𝑅 prime of seven is equal to negative three.

Finally, let’s consider part iii, in which we’re asked to evaluate the integral from one to two of 𝑓 double prime of three 𝑥 with respect to 𝑥. We’re being asked to evaluate a definite integral of the second derivative of the function 𝑓. And in order to do this, we need to recall the second part of the fundamental theorem of calculus. This tells us that if capital 𝐹 is the antiderivative of lower 𝑓 on a particular interval 𝑎𝑏, then the integral from 𝑎 to 𝑏 of lowercase 𝑓 of 𝑡 with respect to 𝑡 is equal to capital 𝐹 of 𝑏 minus capital 𝐹 of 𝑎.

As the first derivative 𝑓 prime of the function is the antiderivative of the second derivative 𝑓 double prime of that function, then we can say that the integral from 𝑎 to 𝑏 of 𝑓 double prime of 𝑡 with respect to 𝑡 is equal to 𝑓 prime of 𝑏 minus 𝑓 prime of 𝑎. This is looking a lot more useful, but we need to consider the variable. Because instead of just 𝑡 or 𝑥, we actually have three 𝑥 as the variable in our second derivative.

We’re going to make a linear substitution; we’ll let 𝑢 equal three 𝑥. d𝑢 by d𝑥, that’s the derivative of 𝑢 with respect to 𝑥, is therefore equal to three. And we can, therefore, say that d𝑥 is equivalent to one-third d𝑢. This integral in 𝑥 would, therefore, be equivalent to the integral of 𝑓 double prime of 𝑢, one-third d𝑢. But what about the limits for this integral? Well, if 𝑢 is equal to three 𝑥, then when 𝑥 is equal to one, 𝑢 is equal to three. So, the lower limit for our new integral in terms of 𝑢 will be three. And when 𝑥 is equal to two, 𝑢 is equal to three times two, that’s six. So, the upper limit for our integral will be six.

We now have an integral in terms of 𝑢 only, the integral from three to six of 𝑓 double prime of 𝑢, one-third d𝑢. Now, this one-third is just a constant, which means we can actually bring it out the front of our integral. That one-third is just a constant multiplier, so we now have one-third the integral from three to six of 𝑓 double prime of 𝑢 with respect to 𝑢. And we’re now able to apply the second part of the fundamental theorem of calculus.

We can say then that this integral is equivalent to one-third 𝑓 prime of six minus 𝑓 prime of three. And both of these values can be found in our table. 𝑓 prime of six is equal to one and 𝑓 prime of three is equal to negative four. So, we have one-third of one minus negative four. That’s one-third of one plus four, which is equal to five over three. By applying the second part of the fundamental theorem of calculus then, we’ve found that the integral from one to two of 𝑓 double prime of three 𝑥 with respect to 𝑥, which is equivalent to our integral in terms of 𝑢, is equal to five-thirds.

We’ve, therefore, completed the problem. In part i, we found that the equation of the line tangent to the composite function ℎ of 𝑥 at 𝑥 equals four is 𝑦 equals 15𝑥 minus 65. In part ii, we found that 𝑅 prime of seven, where 𝑅 is the quotient 𝑔 of 𝑥 over 𝑓 of 𝑥, is negative three. And in part iii, we used the second part of the fundamental theorem of calculus to find the integral from one to two of 𝑓 double prime of three 𝑥 with respect to 𝑥 is five over three.