The Mean Value Theorem and Its Consequences

Read this section to learn about the Mean Value Theorem and its consequences. Work through practice problems 1-3.

Practice Answers

Practice 1: \mathrm{f}^{\prime}(\mathrm{x})=0 when \mathrm{x}=2 and 6 so \mathrm{c}=2 and \mathrm{c}=6.

\mathrm{g}^{\prime}(\mathrm{x})=0 when \mathrm{x}=2, 4, and 6 so \mathrm{c}=2, \mathrm{c}=4, and \mathrm{c}=6.


Practice 2: f(x)=5 x^{2}-4 x+3 on [1,3]. f(1)=4 and f(3)=36 so

\mathrm{m}=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}=\frac{36-4}{3-1}=16

\mathrm{f}^{\prime}(\mathrm{x})=10 \mathrm{x}-4 so \mathrm{f}^{\prime}(\mathrm{c})=10 \mathrm{c}-4=16 if 10 \mathrm{c}=20 and \mathbf{c}=\mathbf{2}.
The graph of \mathrm{f} and the location of \mathrm{c} are shown in Fig. 16.

Fig. 16

Practice 3: If two cars have the same velocities during an interval of time (\mathrm{f}^{\prime}(\mathrm{t})=\mathrm{g}^{\prime}(\mathrm{t}) for \mathrm{t} in \mathrm{I}) then the cars are always a constant distance apart during that time interval.

(Note: The "same velocity" means same speed and same direction. If two cars are traveling at the same speed but in different directions, then the distance between them changes and is not constant)