Question

Tangents Suppose that \(u=g(x)\) is differentiable at \(x=1\) and that \(y=f(u)\) is differentiable at \(u=g(1) .\) If the graph of \(y=f(g(x))\) has a horizontal tangent at \(x=1,\) can we conclude anything about the tangent to the graph of \(g\) at \(x=1\) or the tangent to the graph of \(f\) at \(u=g(1) ?\) Give reasons for your answer.

Short answer

Yes, we can conclude that either the function \(y = f(u)\) at \(u = g(1)\) has a horizontal tangent, or the function \(y = g(x)\) at \(x = 1\) has a horizontal tangent, or both.

Step-by-step solution

Understanding the problem

Given \(y = f(g(x))\) has a horizontal tangent at \(x = 1\). This means that the derivative of \(y = f(g(x))\) at \(x = 1\) is zero. Let's denote \(g(1) = u\). We are asked to explore whether the derivatives of \(y = f(u)\) at \(u = g(1)\), \(y = g(x)\) at \(x = 1\) have any relationships to the above condition or not.

Applying the chain rule for derivatives

We start by applying the chain rule to the function \(y = f(g(x))\). The chain rule gives the derivative of a composition of functions and states that \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). Let's apply this rule at \(x = 1\).

Exploring the consequences

So, by using the chain rule, we have \((f(g(1)))' = f'(g(1)) \cdot g'(1) = f'(u) \cdot g'(1)\). But we know that the tangent to \(f(g(x))\) at \(x = 1\) is horizontal which means \((f(g(1)))' = 0\). So we have \(0 = f'(u) \cdot g'(1)\). This tells us that either \(f'(u) = 0\) or \(g'(1) = 0\) or both. This means that either the function \(y = f(u)\) at \(u = g(1)\) has a horizontal tangent, or the function \(y = g(x)\) at \(x = 1\) has a horizontal tangent, or both.