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GO Classes Test Series 2023 | Calculus | Test 1 | Question: 8

in Calculus edited by
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Choose the CORRECT statement -
  1. The function $f(x)=\exp \left(-x^{2}\right)-1$ has the root $x=0$.
  2. If a function $f$ is differentiable on $[-1,1]$, then there is a point $x$ in that interval where $f^{\prime}(x)=0$.
  3. If $1$ is a root of $f$, then $f^{\prime}(x)$ changes sign at $1 .$
  4. If $f^{\prime \prime}(0)<0$ and $f^{\prime \prime}(1)>0$ then there is a point in $(0,1)$, where $f$ has an inflection point.
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For option D, I think this could be a possible explaination:
Point of inflections are the points of the curve where function changes its curvature (i.e. from concave to convex or vice versa.)
Assuming f'(0)=0 & f'(1)=0, since f''(0)<0 (indicating maxima or convave curvature) & f''(1)>0 (indicating minima or convex curvature) signifies a change in curvature in (0,1). Hence there exists a point of inflection in (0,1).

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The most important note about aboute inflection point is, first of all the function $f(x)$ has to continuous at the inflection point.

Even though you have assumed $f'(0) = f'(1) =0$ it's not enough to conclude that f(x) has an inflection point because what if f''(x) is not even defined.
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@kritz,

Definition 1: A point $(x_0 , f(x_0)) $ where the function graph of a function has a tangent line( no problem if it vertical tangent line also) and where the concavity changes is a Point of Inflection.

$ $

Definition 2:- A point $(x_0 , f(x_0)) $ where the function f(x) is continuous at $x_0$ and where the concavity changes is a Point of Inflection.

In both definitions, The function $f(x)$ was at least continuous at Inflection point and concavity changes.

consider this example

$f(x) =\left\{\begin{matrix} \large x^2, & x < \frac{1}{2}\\ \large{6x},&\frac{1}{2} \leq x \leq \frac{3}{4}\\ \large{-x^2+2x},&x>\frac{3}{4}\end{matrix}\right.$

$f(x)$ is not continuous at $x=\frac{1}{2} , \frac{3}{4}$, $f(x)$  is not differentiable at these points

$ $

$f'(x) =\left\{\begin{matrix} \large 2x, & x < \frac{1}{2}\\ \large{6},&\frac{1}{2} < x < \frac{3}{4}\\ \large{-2x+2},&x>\frac{3}{4}\end{matrix}\right.$

Observe here $f'(0)=0 $ & $f'(1) = 0 $ and  $f'(x)$ is not differentiable at $x=\frac{1}{2} , \frac{3}{4}$

$ $

$f''(x) =\left\{\begin{matrix} \large 2, & x < \frac{1}{2}\\ \large{0},&\frac{1}{2} < x < \frac{3}{4}\\ \large{-2},&x>\frac{3}{4}\end{matrix}\right.$

At $x= \frac{1}{2}$ the function not even continuous moreover the concavity also doesn't change because in interval $(\frac{1}{2},\frac{3}{4})$ f(x) is just a linear. the same happens at $x=\frac{3}{4}$ so there were no inflection points in interval $(0,1)$.

Ex 2:-

$f(x) =\left\{\begin{matrix} \large x^2, & x < \frac{1}{2}\\ \large{5x^2},&\frac{1}{2} \leq x \leq \frac{3}{4}\\ \large{-x^2+2x},&x>\frac{3}{4}\end{matrix}\right.$

$f(x)$ is not continuous at $x = \frac{1}{2} , \frac{3}{4}$, so $f(x)$ is not differentiable at these points.

$ $

$f'(x) =\left\{\begin{matrix} \large 2x, & x < \frac{1}{2}\\ \large{10x},&\frac{1}{2} < x < \frac{3}{4}\\ \large{-2x+2},&x>\frac{3}{4}\end{matrix}\right.$

$  $

$f''(x) =\left\{\begin{matrix} \large 2, & x < \frac{1}{2}\\ \large{10},&\frac{1}{2} < x < \frac{3}{4}\\ \large{-2},&x>\frac{3}{4}\end{matrix}\right.$

You may say that at $x=\frac{3}{4}$ the concavity changes so it will be inflection point, but its not because as per definition $f(x)$ has to be at least continuous at inflection points, but our $f(x)$ is not continuous at $x=\frac{3}{4}$, so it can't be inflection point.

Option D is FALSE.

$ $

Option D Would have been true, if they mention $f''(x)$ exists $\forall x \in (0,1).$

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2 Answers

4 votes
4 votes

  1. The exp function is $1$ at $x = 0.$
  2. Take $f(x) = x$
  3. It is a point, where the second derivative change sign.
  4. By the intermediate number theorem.
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For Option D, IVT is only applicable if $f''(x)$ is continuous in $(0,1)$. So shouldn’t option D be incorrect?
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Yes , Complete context for option D is missing.
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@IchVonViralto A differentiable function is always continuous, therefore option D is true.

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2 votes
2 votes

Option 1:- 

Given ,$f(x)=e^{-x^{2}}-1$

Now , $x=0$ , $f(x)=e^{0}-1=1-1=0$ So , $f(x)$ has a root at $x=0$ .

Option 2:-

The counter example given by the goclass answer will works as $f(x)=x$ .

Option 3:-

Let  $f(x)=x-1$ , So ,$f(x)$ has root at $x=1$ .

Now , $f’(x)=1$ it doesn’t change sign at $x=1$ it is constant function. So it is also false .

Option 4:-

Inflection point :- It is a point in function where the function changes from being concave to convex or vice versa .

Mathematically we can write at inflection point $f’’(x)=0$ .

It is given ,$f''(0)< 0$ and $f''(1)> 0$  . Let ,if we consider $f’’(x)=g(x)$

So , $g(0) <0$ and $g(1)>0$ So we can clearly see the sign of $g(x)$ is change between $[0,1]$ . Now as $f’’(x)$ is exists which implies $g(x)$ is continuous as $f’’(x)=g(x)$ .

So , $g(x)$ is continuous in $[0,1]$ and sign of $g(x)$  change from positive to negative which implies $g(x)$ has a root in $(0,1)$ .

So say at $x_{0}$ is the root of $g(x)$.

So, $g(x_{0})=0$

So,$f’’(x_{0})=0$ 

So , $f(x)$ has an inflection point at $x=x_{0}$ . So option $(D)$ is true .

So correct option is $A$ and $D$.

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thanks for the explanation of option D!!
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My doubt is, how you are concluding $f(x)$ is twice differentiable in interval $(0,1)$?
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