A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Since \(x+7\not= 0\) for all \(x\in [-1,2]\) it follows that the function \(g\text{,}\) as a rational function, is continuous on the closed interval \([-1,2]\) and differentiable on the open interval \((-1,2)\text{.}\) Therefore the function \(g\) satisfies he hypothesis of the Mean Value Theorem on the interval \([-1,2]\text{.}\) By the Mean Value Theorem there is \(c\in (-1,2)\) such that \(\ds g'(c)=\frac{g(2)-g(-1)}{2-(-1)}\text{.}\) Thus the question is to solve \(\ds \frac{21}{(c+7)^2}=\frac{7}{18}\) for \(c\text{.}\) Hence \(\ds c=-7\pm 3\sqrt{6}\text{.}\) Clearly \(-7-3\sqrt{6}\lt -1\) and this value is rejected. From

it follows that \(c=-7+3\sqrt{6}\in(-1,2)\) and it is the only value that satisfies the conclusion of the Mean Value Theorem.

The inequality is obviously satisfied if \(a=b\text{.}\) Let \(a,b\in \mathbb{R}\text{,}\) \(a\lt b\text{,}\) and let \(f(x)=\sin x\text{,}\) \(x\in [a,b]\text{.}\) Clearly the function \(f\) is continuous on the closed interval \([a,b]\) and differentiable on \((a,b)\text{.}\) Thus, by the Mean value Theorem, there is \(c\in (a,b)\) such that \(\ds \cos c=\frac{\sin b-\sin a}{b-a}\text{.}\) Since \(|\cos c|\leq 1\) for all real numbers \(c\) it follows that \(|\sin b-\sin a|\leq |b-a|\text{.}\)

Let \(f(t)\) be the distance that the first horse covers from the start in time \(t\) and let \(g(t)\) be the distance that the second horse covers from the start in time \(t\text{.}\) Let \(T\) be time in which the two horses finish the race. It is given that \(f(0)=g(0)\) and \(f(T)=g(T)\text{.}\) Let \(F(t)=f(t)-g(t)\text{,}\) \(t\in [0,T]\text{.}\) As the difference of two position functions, the function \(F\) is continuous on the closed interval \([0,T]\) and differentiable on the open interval \((0,T)\text{.}\) By the Mean value Theorem there is \(c\in (0,T)\) such that \(\ds F'(c)=\frac{F(T)-F(0)}{T-0}=0\text{.}\) It follows that \(f'(c)=g'(c)\) which is the same as to say that at the instant \(c\) the two horse have the same speed. (Note: It is also possible to use Rolle's theorem.)