Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Using Rolle’s theorem, find the point on the curve y = x(x - 4), where the tangent is parallel to x-axis.

Using Rolle’s theorem, find the point on the curve y = x(x - 4), x\in [0,4] where the tangent is parallel to x-axis.

Answers (1)

Given: y = x(x - 4)

y = (x^2 - 4x)

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

On expanding y=x(x-4), we get  y=\left(x^{2}-4 x\right)
since, x^{2}-4 x is a polynomial and we know that, every polynomial function is continuous for all X\in R
-y=\left(x^{2}-4 x\right)$ is continuous at $x \mathbb{E}[0,4]
Hence, condition 1 is satisfied.
Condition 2
\\ y=\left(x^{2}-4 x\right) \\y^{\prime}=2 x-4
-x^{2}-4 x is differentiable at [0,4]
Hence, condition 2 is satisfied.

Condition 3:

y=x^{2}-4 x \\ x=0 \Rightarrow y=0 \\x=4 \Rightarrow y=(4)^{2}-4(4)=16-16=0

Hence, condition 3 is also satisfied.

Now, there is atleast one value of c ∈ (0,4)

Given tangent to the curve is parallel to the x - axis

This means, Slope of tangent = Slope of x - axis

\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ &2 x-4=0\\ &2 x=4\\ &x=2 \in(0,4)\\ &\text { Put } x=2 \text { in } y=x^{2}-4 x, \text { we have }\\ &y=(2)^{2}-4(2)=4-8=-4 \end{aligned}

Hence, the tangent to the curve is parallel to the x -axis at (2, -4).

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question