One fundamental result about a function that is both continuous on \([a,b]\) and differentiable on \((a,b)\), is the Mean Value Theorem:
If \(f\) is a continuous function on \([a,b]\) and is differentiable on \((a,b)\), then there exists at least one value \(x=c\) in \((a,b)\) where:
\(\displaystyle{\frac{f(b)-f(a)}{b-a}=f^\prime(c)}\)
The Mean Value Theorem is an existence theorem. Meaning, it guarantees the existence of such a point, but there is no guideance on how to find it. There is also a very strong graphing interpretation to the equation in the MVT. The left side of the equation is the calculation for the slope of the secant line that passes through the endpoints of the interval. The calculation on the right side of the equation is the slope of the tangent line at \(x=c\). Anytime slopes are equal, the two lines formed are parallel. Hence the MVT guarantees a value where the tangent line is parallel to the secant line.
Determine whether the MVT applies to the function \(\displaystyle{f(x)=x+\frac{1}{x}}\) on the interval \([1,3]\). If it does, find the point(s) guaranteed to exist. Provide a graph showing the point(s) where the secant line is parallel to the tangent line.
| The only value where \(f\) is not continuous is at \(x=0\) which is not in \([1,3]\). | Similarly, \(f\) is differentiable everywhere but \(x=0\). So the MVT conditions are satified and the MVT can be applied to \(f\). |
| Find the secant line slope: | \(\displaystyle{\begin{array}{l} m_{sec}=\frac{f(3)-f(1)}{3-1} \\ \implies \frac{\left(3+\frac{1}{3}\right)-\left(1+\frac{1}{1}\right)}{2} \\ \implies \frac{\frac{10}{3}-2}{2} \\ \implies \frac{\frac{4}{3}}{2} \\ \implies \frac{2}{3} \end{array}}\) |
| Find the derivative at \(x=c\): | \(\displaystyle{\begin{array}{l} f^\prime(x)=1-\frac{1}{x^2} \\ \implies f^\prime(c)=1-\frac{1}{c^2} \end{array}}\) |
| Set secant line slope equal to \(f^\prime(c)\) and solve for \(c\): | \(\displaystyle{\begin{array}{l} 1-\frac{1}{c^2}=\frac{2}{3} \\ \implies 1-\frac{2}{3}=\frac{1}{c^2} \\ \implies \frac{1}{3}=\frac{1}{c^2} \\ \implies c^2=3 \\ \implies c=\pm\sqrt{3}\end{array}}\) |
| There is only one value at \(c=\sqrt{3}\) on the interval \([1,3]\). | |