Video: Finding the Points on a Curve Where the Slope Has a Given Value Using the Quotient Rule

The function 𝑔 is defined by 𝑔(π‘₯) = π‘₯/(2π‘₯ βˆ’ 1). What points (π‘₯, 𝑦) on the graph of 𝑔 have the property that the line tangent to 𝑔 at (π‘₯, 𝑦) has slope βˆ’1?

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Video Transcript

The function 𝑔 is defined by 𝑔 of π‘₯ equals π‘₯ over two π‘₯ minus one. What points π‘₯, 𝑦 on the graph of 𝑔 have the property that the line tangent to 𝑔 at π‘₯, 𝑦 has slope negative one?

What does it mean for a line to be tangent to the graph of a function at a particular point? Well, it means that that line has the same slope as the graph of the function itself at that point. So in this case, that’s at the point, or possible points, with coordinates π‘₯, 𝑦. We know that the slope of a function is given by its first derivative, in this case, 𝑔 prime of π‘₯. So what this question is really asking us is at what points π‘₯, 𝑦 is 𝑔 prime of π‘₯ equal to negative one.

We need to begin then by finding an expression for 𝑔 prime of π‘₯. And looking at the function 𝑔 of π‘₯, we can see that it is the quotient of two differentiable functions. So in order to find this derivative, we’re going to need to apply the quotient rule. The quotient rule tells us that, for two differentiable functions 𝑒 and 𝑣, the derivative with respect to π‘₯ of their quotient, 𝑒 over 𝑣, is equal to 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared.

Now, our functions in the numerator and denominator are each polynomials. And therefore, they are each differentiable. So we can go ahead and apply the quotient rule. We’ll define 𝑒 to be the function in the numerator β€” that’s π‘₯ β€” and 𝑣 to be the function in the denominator β€” that’s two π‘₯ minus one. We then need to find each of their individual derivatives with respect to π‘₯, which we can do using the power rule of differentiation. The derivative of π‘₯ with respect to π‘₯ is one. And the derivative of two π‘₯ minus one with respect to π‘₯ is two. Remember the derivative of a constant, in this case, negative one, is simply zero.

Substituting into the quotient rule then, we have that 𝑔 prime of π‘₯ is equal to 𝑣 times d𝑒 by dπ‘₯ β€” that’s two π‘₯ minus one times one β€” minus 𝑒 times d𝑣 by dπ‘₯ β€” that’s π‘₯ times two β€” all over 𝑣 squared. That’s two π‘₯ minus one all squared. In the numerator, we have two π‘₯ minus one minus two π‘₯, which simplifies to negative one. So our expression for 𝑔 prime of π‘₯ is negative one over two π‘₯ minus one squared.

Remember, we’re looking for the points where the line tangent to the graph of 𝑔, and hence the graph of 𝑔 itself, has a slope of negative one. So the next step is to set our expression for 𝑔 prime of π‘₯ equal to negative one and solve the resulting equation. We can multiply through by two π‘₯ minus one squared and then divide both sides of the equation by negative one to give one equals two π‘₯ minus one squared.

The next step is to take the square root of each side of this equation. And we must remember to take plus or minus the square root. Plus or minus the square root of one is simply plus or minus one. And the square root of two π‘₯ minus one squared is two π‘₯ minus one. So we have two π‘₯ minus one equals plus or minus one, an equation with two solutions. Either two π‘₯ minus one equals one or two π‘₯ minus one equals negative one.

We can then solve each of these linear equations. For the first equation, we first add one to each side, giving two π‘₯ equals two, and then divide through by two, giving π‘₯ equals one. To solve the second equation, we first add one to each side, giving two π‘₯ is equal to zero, and then divide by two giving π‘₯ equals zero. So we find there are two π‘₯-values at which 𝑔 prime of π‘₯ is equal to negative one, π‘₯ equals zero and π‘₯ equals one.

Remember, though, that we weren’t just asked for the π‘₯-values. We were asked for the coordinates of these points. So finally, we need to evaluate the function itself at each of these π‘₯-values. Substituting into the original function then, we have that 𝑔 of zero is equal to zero over two times zero minus one, which is simply zero. 𝑔 of one is one over two times one minus one. That’s one over two minus one or one over one, which is equal to one. So we found that there are two points π‘₯, 𝑦 at which the line tangent to the graph of 𝑔 has slope negative one. They’re the points zero, zero and one, one.

Remember, in this question, we used the key fact that the line tangent to the graph of a function has the same slope as the function itself at that point. And then, we used the quotient rule in order to find a general expression for our derivative 𝑔 prime of π‘₯. We then set this derivative equal to negative one, solved the resulting equation to find the π‘₯-coordinates of these points, and then used the equation of the original function to find the values of the function itself at these points.

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