Question:

For the function $ f(x) = \frac {4}{3} x^3-8x^2+16x+5, $ $ x = 2\,is \,a\,point\, of $

Updated On: Jun 14, 2022
  • local maxima
  • local minima
  • point of inflexion
  • None of these
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

$f \left(x\right)=\frac{4}{3} x^{3}-8x^{2}+16x+5 \ldots\left(i\right)$
There will be a point of inflexion at point $x$
if $\left(\frac{d^{2}y}{dx^{2}}\right)_{at \left(x=2\right)} =0$, but $\frac{d^{3}y}{dx^{3}} \ne0 $
From E $\left(i\right)$, $f'\left(x\right)=4x^{2}-16x+16$
$f''\left(x\right)=8x-16$
$f''\left(2\right)=16-16=0$
and $f'' \left(x\right)=8\ne0$
Hence, at $x = 2, f\left(x \right)$ shown point of inflexion
Was this answer helpful?
0
0

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