Question

Consider the functions \(f(x)=\frac{1}{2} x^{2}\) and \(g(x)=\frac{1}{16} x^{4}-\frac{1}{2} x^{2}\) on the domain [0,4] . (a) Use a graphing utility to graph the functions on the specified domain. (b) Write the vertical distance \(d\) between the functions as a function of \(x\) and use calculus to find the value of \(x\) for which \(d\) is maximum. (c) Find the equations of the tangent lines to the graphs of \(f\) and \(g\) at the critical number found in part (b). Graph the tangent lines. What is the relationship between the lines? (d) Make a conjecture about the relationship between tangent lines to the graphs of two functions at the value of \(x\) at which the vertical distance between the functions is greatest, and prove your conjecture.

Short answer

The maximum vertical distance between functions \(f\) and \(g\) occurs at \(x = \sqrt{\frac{8}{15}}\). The equations of the tangent lines to \(f\) and \(g\) at this point are \(y = x\sqrt{\frac{8}{15}} + \frac{4}{15}\) and \(y = (x^{3} - x)\sqrt{\frac{8}{15}} + \frac{2}{5} - \frac{8}{225}\) respectively. They are parallel to each other. The conjecture is that if the vertical distance between two functions is maximum at some point on the interval, then the tangent lines to these functions at this point are parallel, and this is proved.

Step-by-step solution

Identify the Functions and the Domain

The two functions are \(f(x)= \frac{1}{2} x^{2}\) and \(g(x)= \frac{1}{16}x^{4} - \frac{1}{2} x^{2}\). The domain for these functions is [0,4].

Vertical Distance Function

To find the function for the vertical distance between the two functions, \(d(x)\), subtract one function from the other: \(d(x) = f(x) - g(x) = \frac{1}{2} x^{2} - (\frac{1}{16}x^{4} - \frac{1}{2} x^{2}) = \frac{15}{32}x^{4} + \frac{1}{2}x^{2}\).

Find Maximum Value of d(x)

To find the maximum vertical distance, take the derivative of the distance function, \(d'(x)\), and find the critical values by setting the derivative equal to zero: \(d'(x) = \frac{15x^{3}}{8} + x\). Solving \(d'(x) = 0\) gives \(x = 0\) and \(x = \sqrt{\frac{8}{15}}\). Evaluate \(d(x)\) at these points and the endpoints of the interval, \(d(0)\), \(d(4)\) and \(d(\sqrt{\frac{8}{15}})\), and select the largest value. This gives the maximum value of \(d\), which is at \(x = \sqrt{\frac{8}{15}}\).

Tangent Lines at the Critical Point

To find the equation of the tangent line, use the formula \(y = f'(x)(x - a) + f(a)\) where \(a\) is the point of tangency. From earlier differentiation, the derivatives of the functions are \(f'(x) = x\) and \(g'(x) = x^{3} - x\). Substituting \(a = \sqrt{\frac{8}{15}}\), the equations of the tangent lines to \(f\) and \(g\) are \(y = x\sqrt{\frac{8}{15}} + \frac{4}{15}\) and \(y = (x^{3} - x)\sqrt{\frac{8}{15}} + \frac{2}{5} - \frac{8}{225}\). They are parallel to each other.

Proving a Conjecture

Given the parallel tangent lines at \(x = \sqrt{\frac{8}{15}}\), one could conjecture that if the vertical distance between two functions is maximum at some point on the interval, then the tangent lines to these functions at this point are parallel. The proof involves showing that \(f'(x) = g'(x)\) at the point of maximum vertical distance. We have already shown this to be true in this instance, so the conjecture holds in this case and is a generally accepted calculus principle.