Maths > Monotonicity, Maxima and Minima > 1.0 Rolle's Theorem
Monotonicity, Maxima and Minima
1.0 Rolle's Theorem
2.0 Lagrange's Mean Value Theorem
3.0 Monotonicity of a function
4.0 Finding intervals of increasing and decreasing functions
5.0 Proving inequality using monotonicity
6.0 Maxima and Minima
7.0 First Derivative Test
8.0 Second Derivative Test
9.0 Absolute Maxima and Absolute Minima
10.0 Maxima and Minima of Discontinuous functions
1.1 Approach to solve the question
Let us first understand the meaning of the terms used in the statement and its significance.
Suppose we are given any function $y=f(x)$ and domain of the function is $[a,b]$. Here closed bracket means both $a$ and $b$ are included or satisfying the function. First of all, we have to check whether we can apply rolle's theorem in the function or not. For that we need to check the conditions given in the statement one by one. Let us check the third condition i.e., $$f(a)=f(b)$$ Put the values of $a$ and $b$ in the function to verify he condition. If it is satisfying then we proceed to check its continuity and differentiability as explained in the previous chapters.
In order to find the value of $'c'$, differentiate the function with respect to $x$ i.e., $$\frac{{dy}}{{dx}} = f'(x)$$Put it equals to $0$ and find the value of $x$ at which $f'(x)=0$ i.e., we get the value of $'c'$.
Note:
1. If we do not get atleast one real value of $x$ after putting $f(x)$ equals to $0$, then rolle's theorem can not be applied.
2. $f'(c)=0$ also signify that at $x=c$, there is a horizontal tangent to the curve of the function $y=f(x)$ as shown in figure.
3. We can also say that between two zeroes/roots of $f(x)$ between $a$ and $b$ (i.e., two values of $x$ at which $f(x)=0$) there exists atleast one zero/root of $f'(x)$.
4. Always remember, all polynomial functions, exponential functions, logarithmic functions are continuous and differentiable. So, there is no need to check all the conditions, directly check the third condition i.e., $f(a)=f(b)$.
5. To check whether the given function $f(x)$ has atleast one real root in the given interval $(a,b)$ just integrate the function in the interval i.e., $$\int\limits_a^b {f(x)dx} $$If the value of the integral is equals to $0$, then there exists atleast one real root in that particular interval.
Question 1. Determine whether Rolle's theorem can be applied , if it can, find all the values of $c$ such that $f'(c)=0$ and $f(x) = ({x^2} - 2x){e^x}$ in the interval $[0,2]$.
Solution: The interval given is $[0,2]$. Since the function $f(x)$ can be written as the product of two functions i.e., $$f(x) = ({x^2} - 2x){e^x} = g(x).h(x)$$ Since both $g(x)$ and $h(x)$ are continuous and differentiable function, $f(x)$ is also continuous and differentiable in $[0,2]$. Now, we check the third condition i.e., $$\begin{equation} \begin{aligned} f(0) = ({0^2} - 2.0){e^0} = 0 \\ f(2) = ({2^2} - 2.2){e^2} = 0 \\\end{aligned} \end{equation} $$We can say that $$f(a) = f(b)$$Now, to find the value of $c$ between $a$ and $b$, we differentiate the function with respect to $x$ i.e., $$f'(x) = (2x - 2){e^x} + ({x^2} - 2x){e^x}$$ Put it equals to $0$, we get $$\begin{equation} \begin{aligned} f'(x) = (2x - 2){e^x} + ({x^2} - 2x){e^x} = 0 \\ {e^x}(2x - 2 + {x^2} - 2x) = 0 \\ {e^x}({x^2} - 2) = 0 \\ \Rightarrow x = \pm \sqrt 2 \\\end{aligned} \end{equation} $$Therefore, we can say that $$f'(\sqrt 2 ) = 0\;{\text{and }}f'( - \sqrt 2 ) = 0$$ But the value of $c$ lies between $[0,2]$, therefore, we get $$c = \sqrt 2 $$
Question 2. Prove that the function $f(x) = \sin x + x\cos x$ has atleast one root in the interval $(0,\pi)$.
Solution: To check the roots in the interval, we have to integrate the function in given interval i.e., $$\begin{equation} \begin{aligned} \int\limits_0^\pi {f(x)} dx = \int\limits_0^\pi {\left( {\sin x + x\cos x} \right)} dx \\ = \int\limits_0^\pi {\sin xdx} + \int\limits_0^\pi {x\cos xdx} \\ = - (\cos \pi - cos0) + x\int\limits_0^\pi {\cos xdx} - \int\limits_0^\pi {\left( {\frac{d}{{dx}}(x).\int {\cos xdx} } \right)} dx \\ = - ( - 1 - 1) + x(\sin \pi - \sin 0) - \int\limits_0^\pi {1.\sin xdx} \\ = 2 + x(0 - 0) + (cos\pi - cos0) \\ = 2 - 2 = 0 \\\end{aligned} \end{equation} $$ Since, the value of the integral is equals to $0$, then there exists atleast one real root in the interval $(0,\pi)$.
Question 3. If $f(x) = {x^a}\log x$ and $f(0)=0$, then the value of $'a'$ for which Rolle's theorem can be applied in $[0,1]$ is
(a) $-2$
(b) $-1$
(c) $0$
(d) $1/2$
Solution: As it is clear from the equation of the function that it is continuous and differentiable in the interval $[0,1]$. For, Rolle's theorem to be applicable $f'(c)=0$ i.e., $$\begin{equation} \begin{aligned} f'(x) = {x^a}\frac{d}{{dx}}(\log x) + \frac{d}{{dx}}({x^a})\log x \\ f'(x) = \frac{{{x^a}}}{x} + a{x^{a - 1}}\log x \\ f'(x) = {x^{a - 1}} + a{x^{a - 1}}\log x \\ f'(x) = {x^{a - 1}}(1 + a\log x) = 0 \\ \Rightarrow 1 + a\log x = 0 \\ \Rightarrow \log x = - \frac{1}{a} \\ \Rightarrow x = {e^{ - \frac{1}{a}}} \\\end{aligned} \end{equation} $$Since the function is $f(x) = {x^a}\log x$, therefore, $$x \ne 0$$ From this, we get $$\begin{equation} \begin{aligned} x = {e^{ - \frac{1}{a}}} \ne 0 \\ \frac{1}{{{e^{\frac{1}{a}}}}} \ne 0 \\ \Rightarrow a \ne 0 \\\end{aligned} \end{equation} $$ Also, there is one condition on $c$ that it should lie between $a$ and $b$. From this condition, we can write $$0 \leqslant {e^{ - \frac{1}{a}}} \leqslant 1$$ Since, ${e^{ - \frac{1}{a}}}$ will always positive, we can write, $$\begin{equation} \begin{aligned} {e^{ - \frac{1}{a}}} \leqslant 1 \\ \log {e^{ - \frac{1}{a}}} \leqslant \log 1 \\ - \frac{1}{a} \leqslant 0 \\ \frac{1}{a} \geqslant 0 \\ \Rightarrow a \geqslant 0 \\\end{aligned} \end{equation} $$ Therefore, the correct answer is option $d$.
