Question:
Find all the points of local maxima and local minima of the function $f(x) = (x - 1)^3 (x + 1)^2$
Find all the points of local maxima and local minima of the function $f(x) = (x - 1)^3 (x + 1)^2$
Updated On: Jul 6, 2022
- $1$, $-1$, $-1/5$
- $1$, $-1$
- $1$, $-1/5$
- $-1$, $-1/5$
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Let $y = f(x) = (x - 1)^3(x + 1)^2$. Then,
$\frac{dy}{dx} = 3\left(x-1\right)^{2} \left(x + 1\right)^{2} + 2\left(x+ 1\right)\left(x - 1\right)^{3}$
$\Rightarrow \frac{dy}{dx} = \left(x - 1\right)^{2 }\left(x + 1\right)\left\{3\left(x +1\right) + 2\left(x -1\right)\right\}$
$\Rightarrow \frac{dy}{dx} = \left( x - 1\right)^{ 2} \left(x+1\right)\left(5x + 1\right)$
For local maximum or local minimum, we have
$ \frac{dy}{dx} = 0 \Rightarrow \left(x -1\right)^{2}\left(x + 1\right)\left(5x + 1\right) = 0$
$\Rightarrow x = 1$ or, $x = - 1$ or,
$x = -\frac{1}{5}$
Was this answer helpful?
0
0
Top Questions on Application of derivatives
- For the function $f(x) = x^3 - 6x^2 + 12x - 3$, $x = 2$ is:}
- KCET - 2024
- Mathematics
- Application of derivatives
- If \( \frac{dy}{dx} - y \log_e 2 = 2^{\sin x} (\cos x - 1) \log_e 2 \), then \( y \) is:
- BITSAT - 2024
- Mathematics
- Application of derivatives
- The solution of the differential equation \( (x + 1)\frac{dy}{dx} - y = e^{3x}(x + 1)^2 \) is:
- BITSAT - 2024
- Mathematics
- Application of derivatives
- If the area bounded by the curves \( y = ax^2 \) and \( x = ay^2 \) (where \( a>0 \)) is 3 sq. units, then the value of \( a \) is:
- BITSAT - 2024
- Mathematics
- Application of derivatives
- The area of the region bounded by the curves \( x = y^2 - 2 \) and \( x = y \) is:
- BITSAT - 2024
- Mathematics
- Application of derivatives
View More Questions
Notes on Application of derivatives
- Applications of Derivatives Mathematics
- NCERT Solutions For Class 12 Mathematics Chapter 6 Applications of Derivatives
- NCERT Solutions for Class 12 Maths Chapter 6 Applications of Derivatives 6 1
- NCERT Solutions for Class 12 Maths Chapter 6 Applications of Derivatives 6 2
- NCERT Solutions for Class 12 Maths Chapter 6 Applications of Derivatives 6 3
- NCERT Solutions for Class 12 Maths Chapter 6 Applications of Derivatives 6 4
Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives