Question

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. Graph the function, the secant line through the endpoints, and the tangent line at . Are the secant line and the tangent line parallel?

14.\(f(x) = {e^{ - x}}\), \(\left( {0\;,\;2} \right)\)

Short answer

The number that satisfies the conclusion of the Mean value Theorem is \(c = \ln \left( {\frac{2}{{1 - {e^{ - 2}}}}} \right)\).

Step-by-step solution

Given data

The function \(f(x) = {e^{ - x}}\) on the given interval, \(\left( {0\;,\;2} \right)\).

Concept of Mean value theorem

Let\(f\)be a function that satisfies the following hypothesis:

1.\(f\)is continuous on the closed interval\(\left( {a\;,{\rm{ }}b} \right)\).

2.\(f\)is differentiable on the open interval\((a\;,{\rm{ }}b)\).

Then, there is a number\(c\)in\(\left( {a\;,{\rm{ }}b} \right)\)such that\({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently,\(f(b) - f(a) = {f^\prime }(c)(b - a)\).

Calculation for the derivative of function \(f(x)\)

Obtain the derivative of\(f(x)\).

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{e^{ - x}}} \right)\\{f^\prime }(x) &= - {e^{ - x}}\end{aligned}\)

Replace\(x\)by\(c\)in\({f^\prime }(x)\).

\({f^\prime }(c) = - {e^{ - c}}\)

If\(a = 0\), then the value of\(f(0)\)is given below.

\(\begin{aligned}{c}f(a) &= f(0)\\ &= {e^{ - 0}}\\ &= 1\end{aligned}\)

If\(b = 2\), then the value of\(f(2)\)is shown below.

\(\begin{aligned}{c}f(b) &= f(2)\\ &= {e^{ - 2}}\end{aligned}\)

Calculation of the value of \(c\)

Compute the values of\(c\)such that\({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Substitute the corresponding values.

\(\begin{aligned}{c} - {e^{ - c}} &= \frac{{{e^{ - 2}} - 1}}{{2 - 0}}\\ - {e^{ - c}} &= \frac{{{e^{ - 2}} - 1}}{2}\\ - {e^{ - c}} &= \frac{{ - \left( {1 - {e^{ - 2}}} \right)}}{2}\\{e^{ - c}} &= \frac{{1 - {e^{ - 2}}}}{2}\end{aligned}\)

Take logarithm on both the sides.

\(\begin{aligned}{c} - c &= \ln \left( {\frac{{1 - {e^{ - 2}}}}{2}} \right)\\c &= - \ln \left( {\frac{{1 - {e^{ - 2}}}}{2}} \right)\\c &= \ln {\left( {\frac{{1 - {e^{ - 2}}}}{2}} \right)^{ - 1}}\end{aligned}\)

Thus, the value of\(c = \ln \left( {\frac{2}{{1 - {e^{ - 2}}}}} \right)\).

The equation of the secant line through the end points is,\(\frac{{y - f(a)}}{{x - a}} = \frac{{f(b) - f(a)}}{{b - a}}\).

\(\begin{aligned}{c}\frac{{y - 1}}{{x - 0}} &= \frac{{{e^{ - 2}} - {e^{ - 0}}}}{{2 - 0}}\\\frac{{y - 1}}{x} &= \frac{{{e^{ - 2}} - 1}}{2}\\y - 1 &= \frac{{{e^{ - 2}} - 1}}{2}x\\y &= \frac{{{e^{ - 2}} - 1}}{2}x + 1\end{aligned}\)

Thus, the required secant line is \(y = \frac{{{e^{ - 2}} - 1}}{2}x + 1\).

Calculation for the tangent function

The equation of the tangent at\((c,f(c))\)is,\({f^\prime }(c) = \frac{{y - f(c)}}{{x - c}}\).

Express the above equation as,\(y = {f^\prime }(c)(x - c) + f(c)\).

\(\begin{aligned}{c}y &= - {e^{ - c}}(x - c) + {e^{ - c}}\\y &= - {e^{ - c}}(x - c - 1)\\y &= - {e^{ - \ln \left( {\frac{2}{{1 - {e^{ - 2}}}}} \right)}}\left( {x - \ln \left( {\frac{2}{{1 - {e^{ - 2}}}}} \right) - 1} \right)\\y &= - {e^{\ln {{\left( {\frac{2}{{1 - {e^{ - 2}}}}} \right)}^{ - 1}}}}\left( {x - \ln \left( {\frac{2}{{1 - {e^{ - 2}}}}} \right) - 1} \right)\end{aligned}\)

On further simplification, the equation becomes:

\(y = \left( {\frac{{{e^{ - 2}} - 1}}{2}} \right)\left( {x - \ln \left( {\frac{2}{{1 - {e^{ - 2}}}}} \right) - 1} \right)\)

Thus, the required tangent line is \(y = \left( {\frac{{{e^{ - 2}} - 1}}{2}} \right)\left( {x - \ln \left( {\frac{2}{{1 - {e^{ - 2}}}}} \right) - 1} \right)\).

Sketch the graph of the secant and tangent lines

Use the graph calculator to draw the graph of the secant and the tangent lines as shown below in Figure (1).

From Figure 1, it is noticed that the secant line and the tangent line is parallel to each other.