Calculus

Applications of the Derivative

Applications of Derivative Logo

Rolle’s Theorem

Solved Problems

Example 9.

Check the validity of the Rolle's theorem for the function \[f\left( x \right) = \frac{{{x^2} - 4x + 3}}{{x - 2}}\] on the segment \(\left[ {1,3} \right].\)

Solution.

It can be seen that the values of the function at the endpoints are equal:

\[f\left( 1 \right) = \frac{{{1^2} - 4 \cdot 1 + 3}}{{1 - 2}} = 0,\;\;f\left( 3 \right) = \frac{{{3^2} - 4 \cdot 3 + 3}}{{3 - 2}} = 0.\]

However, when \(x = 2\), the function has a discontinuity, i.e. one of the three conditions of Rolle's theorem − the condition of continuity of the function on the interval \(\left[ {a,b} \right]\) − is not satisfied. Therefore, in this case, Rolle's theorem does not apply.

Example 10.

Check the validity of Rolle's theorem for the function \[f\left( x \right) = \left| {x - 1} \right|\] on the segment \(\left[ {0,2} \right].\)

Solution.

This function is continuous on the interval \(\left[ {0,2} \right]\) and has equal values at the endpoints of the interval:

\[f\left( 0 \right) = \left| {0 - 1} \right| = \left| { - 1} \right| = 1,\;\;f\left( 2 \right) = \left| {2 - 1} \right| = \left| 1 \right| = 1.\]

However, the third condition of Rolle's theorem − the requirement for the function being differentiable on the open interval \(\left( {0,2} \right)\) − is not satisfied, because the derivative does not exist at \(x = 1\) (the function has a cusp at this point). Thus, in this case, Rolle's theorem can not be applied.

Example 11.

Let \[f\left( x \right) = {x^3} - 3x.\] Find all values of \(c\) in the interval \(\left( { - 1,2} \right)\) such that \(f^\prime\left( c \right) = 0.\)

Solution.

Graph of the function f(x)=x^3-3x
Figure 8.

First we check that the function \(f\left( x \right)\) satisfies all the conditions of Rolle's theorem.

\(1.\) \(f\left( x \right)\) is a cubic function. Hence, it is continuous everywhere and, in particular, over the closed interval \(\left[ { - 1,2} \right].\)

\(2.\) \(f\left( x \right)\) is also differentiable everywhere as a polynomial function.

\(3.\) Notice that

\[f\left( { - 1} \right) = {\left( { - 1} \right)^3} - 3 \cdot \left( { - 1} \right) = 2,\]
\[f\left( 2 \right) = {2^3} - 3 \cdot 2 = 2,\]

that is \(f\left( { - 1} \right) = f\left( 2 \right).\)

So we may apply Rolle's theorem, which guarantees that there is at least one point \(c\) in the open interval \(\left( { - 1,2} \right)\) where \(f^\prime\left( c \right) = 0.\)

Solve the equation:

\[f'\left( c \right) = 0,\;\;\Rightarrow 3{c^2} - 3 = 0,\;\; \Rightarrow 3\left( {{c^2} - 1} \right) = 0,\;\; \Rightarrow 3\left( {c - 1} \right)\left( {c + 1} \right) = 0,\;\; \Rightarrow {c_1} = - 1,\;{c_2} = 1.\]

We see that only one of the roots belongs to the open interval \(\left( { - 1,2} \right).\) So the answer is \(c = 1.\)

Example 12.

Given the function \[f\left( x \right) = {x^3} - 4x + 1.\] Find all values of \(c\) in the interval \(\left( {-2,2} \right)\) such that \(f^\prime\left( c \right) = 0.\)

Solution.

First we make sure that we can apply Rolle's theorem. As we deal with a polynomial, this function is continuous and differentiable in the given interval. Compute the values at the endpoints of the interval:

\[f\left( { - 2} \right) = {\left( { - 2} \right)^3} - 4 \cdot \left( { - 2} \right) + 1 = 1,\]
\[f\left( 2 \right) = {2^3} - 4 \cdot 2 + 1 = 1.\]

Thus, the function has equal values at the endpoints. Hence, all three conditions of Rolle's theorem hold.

Find the values of \(c.\)

\[f^\prime\left( x \right) = \left( {{x^3} - 4x + 1} \right)^\prime = 3{x^2} - 4.\]
\[f^\prime\left( c \right) = 0,\;\; \Rightarrow 3{c^2} - 4 = 0,\;\; \Rightarrow 3{c^2} = 4,\;\; \Rightarrow {c^2} = \frac{4}{3},\;\; \Rightarrow {c_{1,2}} = \pm \frac{2}{{\sqrt 3 }} = \pm \frac{{2\sqrt 3 }}{3} \approx \pm 1.15\]

So we obtained two values \({c_1} = - \frac{{2\sqrt 3 }}{3}\) and \({c_2} = \frac{{2\sqrt 3 }}{3}\) where the derivative of the function is zero. Both of them belong to the open interval \(\left( {-2,2} \right).\)

Example 13.

Let \[f\left( x \right) = \ln \left( {2 - {x^2}} \right).\] Find all values of \(c\) in the open interval \(\left( { - 1,1} \right)\) such that \(f^\prime\left( c \right) = 0.\)

Solution.

First, we determine the domain of the function:

\[2 - {x^2} \gt 0,\;\; \Rightarrow D = \left( { - \sqrt 2 ,\sqrt 2 } \right).\]

Since the interval \(\left( { - 1,1} \right)\) belongs the domain of the function, the function is continuous and differentiable on \(\left( { - 1,1} \right).\)

The function is even, so \(f\left( { - 1} \right) = f\left( 1 \right).\)

Hence, Rolle's theorem is applicable to this function.

We find the derivative by the chain rule:

\[f^\prime\left( x \right) = \left( {\ln \left( {2 - {x^2}} \right)} \right)^\prime = \frac{1}{{2 - {x^2}}} \cdot \left( {2 - {x^2}} \right)^\prime = \frac{{ - 2x}}{{2 - {x^2}}}.\]

Equating the derivative to zero, we get the values of \(c:\)

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow \frac{{ - 2c}}{{2 - {c^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { - 2c = 0}\\ {2 - {c^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow c = 0.\]

Example 14.

Determine the number of stationary points of the function \[f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)\] and indicate the intervals, in which they are located.

Solution.

The function has zeros at the following points:

\[{x_1} = 0,\;\;{x_2} = 1,\;\;{x_3} = 2.\]

Consider two segments: \(\left[ {0,1} \right]\) and \(\left[ {1,2} \right].\) Obviously, the conditions of Rolle's theorem are satisfied on each of these segments. Therefore, there are two stationary points \({\xi_1}\) and \({\xi_2},\) lying in the following intervals:

\[{\xi _1} \in \left( {0,1} \right),\;\;{\xi _2} \in \left( {1,2} \right).\]

To verify this conclusion, we find the derivative and calculate the stationary points.

\[f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right) = \left( {{x^2} - x} \right)\left( {x - 2} \right) = {x^3} - {x^2} - 2{x^2} + 2x = {x^3} - 3{x^2} + 2x;\]
\[f'\left( x \right) = {\left( {{x^3} - 3{x^2} + 2x} \right)^\prime } = 3{x^2} - 6x + 2.\]

The roots of the derivative are

\[f'\left( x \right) = 0,\;\; \Rightarrow 3{x^2} - 6x + 2 = 0,\;\; \Rightarrow D = 36 - 4 \cdot 3 \cdot 2 = 12,\;\; \Rightarrow {x_{1,2}} = \frac{{6 \pm \sqrt {12} }}{6} = 1 \pm \frac{1}{{\sqrt 3 }} \approx 0,42;\;1,58.\]

So, the first stationary point \({x_1} \approx 0,42\) lies in the interval \(\left( {0,1} \right),\) and the second point \({x_2} \approx 1,58\) lies in the interval \(\left( {1,2} \right),\) which coincides with the above results.

Example 15.

Determine whether the function \[f\left( x \right) = \sin \frac{x}{2}\] satisfies conditions of Rolle's theorem for the interval \(\left[ {0,2\pi } \right].\) If so find all numbers \(c\) that satisfy the conclusion of the theorem.

Solution.

The function is continuous on \(\left[ {0,2\pi } \right]\) and differentiable on the interval \(\left( {0,2\pi } \right).\) Check that the function has equal values at the endpoints of the interval:

\[f\left( 0 \right) = \sin \frac{0}{2} = 0,\]
\[f\left( {2\pi } \right) = \sin \frac{{2\pi }}{2} = \sin \pi = 0.\]

Hence, we can apply Rolle's theorem to this function.

Find the derivative:

\[f^\prime\left( x \right) = \left( {\sin \frac{x}{2}} \right)^\prime = \frac{1}{2}\cos \frac{x}{2}.\]

Solve the trig equation to determine the values of \(c.\)

\[f^\prime\left( c \right) = 0,\;\; \Rightarrow \frac{1}{2}\cos \frac{c}{2} = 0,\;\; \Rightarrow \frac{c}{2} = \frac{\pi }{2} + \pi n,\;\; \Rightarrow c = \pi + 2\pi n,\,n \in \mathbb{Z}.\]

It is easy to see that only one point \(c = \pi\) belongs to the interval \(\left( {0,2\pi } \right)\). So the answer is \(c = \pi.\)

Example 16.

Check the validity of Rolle's theorem for the function \[ f(x) = \begin{cases} x^2, & \text{if}\;\;\; 0 \le x \le 2 \\ 6-x, & \text{if}\;\;\; 2 \lt x \le 6 \end{cases}. \]

Solution.

The function is defined and continuous on the interval \(\left[ {0,6} \right]\) and takes equal values at the endpoints of the segment:

\[f\left( 0 \right) = {0^2} = 0,\;\;f\left( 6 \right) = 6 - 6 = 0.\]

However, this function is not differentiable at \(x = 2.\) Indeed, when calculating the derivative at this point, the result depends on the sign of the increment \(\Delta x:\)

If \(\Delta x \lt 0,\) we have

\[f'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{{{\left( {x + \Delta x} \right)}^2} - {x^2}}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\cancel{x^2} + 2x\Delta x + {{\left( {\Delta x} \right)}^2} - \cancel{x^2}}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \left( {2x + \Delta x} \right) = 2x,\;\;\Rightarrow f'\left( {2 - 0} \right) = 4. \]

For \(\Delta x \gt 0\) we obtain:

\[f'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\left[ {6 - \left( {x + \Delta x} \right)} \right] - \left( {6 - x} \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\cancel{6} - \cancel{x} - \Delta x - \cancel{6} + \cancel{x}}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \left( { - \frac{{\cancel{\Delta x}}}{{\cancel{\Delta x}}}} \right) = - 1,\;\; \Rightarrow f'\left( {2 + 0} \right) = - 1.\]

This means that the derivative does not exist at the point \(x = 2\) (Figure \(9\)), that is, the condition of differentiability of the function at all points of the interval is not satisfied.

Application of Rolle's theorem to a piecewise function
Figure 9.

Therefore, Rolle's theorem is inapplicable here.

Example 17.

Determine whether the function \[f\left( x \right) = \text{sech}\,x = \frac{2}{{{e^x} + {e^{ - x}}}}\] satisfies conditions of Rolle's theorem for the interval \(\left[ {-1,1 } \right].\) If so find all numbers \(c\) that satisfy the conclusion of the theorem.

Solution.

The hyperbolic secant function \(f\left( x \right) = \text{sech}\,x = \frac{2}{{{e^x} + {e^{ - x}}}}\) is continuous and differentiable for all real \(x\). As this is an even function, it is obvious that \(f\left( { - 1} \right) = f\left( 1 \right).\) Therefore Rolle's theorem is applicable to this function.

Find the derivative and equate it to zero to get \(c.\)

\[f^\prime\left( x \right) = {\left( {\text{sech}\,x}\right)}^\prime = \left( {\frac{2}{{{e^x} + {e^{ - x}}}}} \right)^\prime = - \frac{2}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cdot \left( {{e^x} + {e^{ - x}}} \right)^\prime = - \frac{{2\left( {{e^x} - {e^{ - x}}} \right)}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}.\]
\[f^\prime\left( c \right) = 0,\;\; \Rightarrow - \frac{{2\left( {{e^c} - {e^{ - c}}} \right)}}{{{{\left( {{e^c} + {e^{ - c}}} \right)}^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2\left( {{e^c} - {e^{ - c}}} \right) = 0}\\ {{{\left( {{e^c} + {e^{ - c}}} \right)}^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow {e^c} - {e^{ - c}} = 0,\;\; \Rightarrow {e^{2c}} = 1,\;\; \Rightarrow {e^{2c}} = {e^0},\;\; \Rightarrow 2c = 0,\;\; \Rightarrow c = 0.\]
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