Question
Verify Rolle's theorem for the following function on the indicated intervals
f(x) = x(x - 4)2 on the interval [0, 4]
✓
Answer
Given function is f(x) = x(x - 4)2. Which can be rewritten as f(x) = x3 - 8x2 + 16x.
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function f(x) is continuous and derivable on [0, 4].
Also,
f(0) = f(4) = 0
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists $\text{c}\in[0,4]$ such that f'(c) = 0.
We have
f(x) = x3 - 8x2 + 16x
⇒ f'(x) = 3x2 - 16x + 16
$\therefore$ f'(x) = 0
⇒ 3x2 - 16x + 16 = 0
⇒ 3x2 - 12x - 4x + 16 = 0
⇒ 3x(x - 4) - 4(x - 4) = 0
⇒ (x - 4)(3x - 4)
$\Rightarrow\text{x}=4,\frac{4}{3}$
Thus, $\text{c}=\frac{4}{3}\in(0,4)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function f(x) is continuous and derivable on [0, 4].
Also,
f(0) = f(4) = 0
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists $\text{c}\in[0,4]$ such that f'(c) = 0.
We have
f(x) = x3 - 8x2 + 16x
⇒ f'(x) = 3x2 - 16x + 16
$\therefore$ f'(x) = 0
⇒ 3x2 - 16x + 16 = 0
⇒ 3x2 - 12x - 4x + 16 = 0
⇒ 3x(x - 4) - 4(x - 4) = 0
⇒ (x - 4)(3x - 4)
$\Rightarrow\text{x}=4,\frac{4}{3}$
Thus, $\text{c}=\frac{4}{3}\in(0,4)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the equation of the plane passing through the intersection of the planes $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})=7,\vec{\text{r}}\cdot(2\hat{\text{i}}+5\hat{\text{j}}+3\hat{\text{k}})=9$ and the point (2, 1, 3).
→Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{k}})+\lambda\hat{\text{i}}+\mu(\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
→$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{k}})+\lambda\hat{\text{i}}+\mu(\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
Evaluate the following integrals:
→$\int\text{e}^{\text{x}}\frac{1+\text{x}}{(2+\text{x})^2}\text{dx}$
If $\text{y}=\cos^{-1}(2\text{x})+2\cos^{-1}\sqrt{1-4\text{x}^2}, -\frac{1}{2}<\text{x}<0,$ find $\frac{\text{dy}}{\text{dx}}.$
→In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}\text{at x} = \pi$
→$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}\text{at x} = \pi$
If $\text{A}=\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix},$ prove that
$ \text{A}^2=\begin{bmatrix}\cos\text{n}\alpha+\sin\text{n}\alpha&\sqrt{2}\sin\text{n}\alpha\\-\sqrt{2}\sin\text{n}\alpha&\cos\text{n}\alpha-\sin\text{n}\alpha\end{bmatrix}$ for all $\text{n}\in\text{N}.$
→$ \text{A}^2=\begin{bmatrix}\cos\text{n}\alpha+\sin\text{n}\alpha&\sqrt{2}\sin\text{n}\alpha\\-\sqrt{2}\sin\text{n}\alpha&\cos\text{n}\alpha-\sin\text{n}\alpha\end{bmatrix}$ for all $\text{n}\in\text{N}.$
Find the equation of the plane which bisects the line segment joining the points (-1, 2, 3) and (3, -5, 6) at right angles.
$\text{A(adj. A) = (adj.A)A} | \text{A} = | \text{I}_{3}.$
If a, b and c are all non-zero and $ \begin{vmatrix} \text{1 + a} & 1 & 1 \\ 1 & \text{1 + b} & 1 \\ 1 & 1 & \text{1 + c} \end{vmatrix} = 0,$ then prove that $\frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} 1 = 0.$
→If a, b and c are all non-zero and $ \begin{vmatrix} \text{1 + a} & 1 & 1 \\ 1 & \text{1 + b} & 1 \\ 1 & 1 & \text{1 + c} \end{vmatrix} = 0,$ then prove that $\frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} 1 = 0.$
Using properties of determinants, prove that
; background:rgba(220,220,220,0.5))
![]( "Click and drag to move")
If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that at $\text{t}=\frac{\pi}{4},\big(\frac{\text{dy}}{\text{dx}}\big)=\frac{\text{b}}{\text{a}}$
→