Question

Let function $ f(x)={{(x-1)}^{2}}{{(x+1)}^{3}}. $ Then, which of the following is false?

• A. There exists a point where $ f(x) $ has a maximum value
• B. There exists a point where $ f(x) $ has a minimum value
• C. There exists a point where $ f(x) $ has neither maximum nor minimum value
• D. All of the above

Answer 1

The Correct Option is D

Given, $ f(x)={{(x-1)}^{2}}{{(x+1)}^{3}} $ $ f'(x)={{(x-1)}^{2}}3{{(x+1)}^{2}}+2(x-1){{(x+1)}^{3}} $
$=(x-1){{(x+1)}^{2}}[3(x-1)+2(x+1)] $
$=(x-1){{(x+1)}^{2}}[3x-3+2x+2] $
$ \Rightarrow $ $ f'(x)=(x-1)\,{{(x+1)}^{2}}\,(5x-1) $ ..(i)
For maxima or minima, $ f'(x)=0 $
$ \Rightarrow $ $ (x-1){{(x+1)}^{2}}(5x-1)=0 $
$ \Rightarrow $ $ x=-1,\,\,\,1,\frac{1}{5} $
Again, differentiating E (i) . r. t, x. we get $ f''(x)=(x-1)(5x-1)\frac{d}{dx}{{(x+1)}^{2}} $ $ +{{(x+1)}^{2}}\frac{d}{dx}(x-1)(5x-1) $
$=(5{{x}^{2}}-6x+1)2(x+1)+{{(x+1)}^{2}}(10x-6) $
$=2[5{{x}^{3}}-{{x}^{2}}-5x+1]+(5{{x}^{3}}+7{{x}^{2}}-x-3)\,(2) $
$ \Rightarrow $ $ f''(x)=2[10{{x}^{3}}+6{{x}^{2}}-6x-2] $ At $ x=-1,\,f''(x)=0 $
At $ x=\frac{1}{5},\,\,f''(x)<0 $ and at $ x=1,\,\,f''(x)>0 $
So, at $ x=-1, $
$ f(x) $ has neither maximum not minimum value At $ x=\frac{1}{5},\,\,f(x) $ has maximum value
At $ x=1,\,\,\,f(x) $ has minimum value

Answer 2

Ans. The idea of derivatives may be used to locate the maximum and minimum of a function. To determine the gradient or slope of a function, derivatives are utilized; the places where the gradient is zero are referred to as turning or stationary points. These locations reveal the function's (locally) highest or lowest values.

Let, f(x) is the real function having an I interval. Then, f(x) will have the maximum value in the interval I if and only if a point “c” in the interval satisfies f(x) ≤ f(c) for all values of x ∈ I.

So, f(c) will be the maximum value function f(x) in an I interval, and point c will be known as the maximum value point of the f function in Interval I.

Let, f(x) is a real function having an interval I. Then f(x) will have the minimum value in the interval I, if there is a point “c” in the interval, satisfy f(x) ≥ f(c) for all values of x ∈ I.

So, f(c) will be the minimum value of f(x) in the given interval I and point c will be known as the minimum value in the given interval I.

f(x) is an extreme value function if a point c in the given interval I such that f(c) is either a maximum value or a minimum value of f(x) in the interval I and the point c is known as an extreme point.