Question

Which of the following function has a point of inflection at x = 0 ?

• A. $f\left(x\right)=sin\left(x\right)$
• B. $f\left(x\right)=cos\left(x\right)$
• C. $f\left(x\right)=x^2$
• D. $f\left(x\right)=x^3$

Answer 1

Answer: (A), (D)

The correct options are
A

$f\left(x\right)=sin\left(x\right)$

D

$f\left(x\right)=x^3$

We will find f”(x) and substitute x=0 to find check if zero is a point of inflection. At point of inflection, we also have $f{”}^{\prime }\left(x\right)\ne 0$

Let’s check each option one by one.

1. f(x) = sin (x) (Given)

f’(x) = cos(x)

f”(x) = - sin (x)

f”(0) = 0

So x = 0 could be a point of inflection.

To confirm it let’s check f”’(x) as well.

”’(x) = - cos(x)

$f{”}^{\prime }\left(0\right)\ne 0$

Hence, we can say x = 0 is a point of inflection for f(x) = sin (x).

b. f(x) = cos(x) (Given)

f’(x) = - sin(x)

f”(x) = - cos(x)

f”(0) = -1

Since, f”(0) ≠ 0 , x = 0 is not a point of inflection.

c. $f\left(x\right)=x^2$

f’’(x) = 2

=> f”(0) = 2

Since f”(0) is not equal to zero, x = 0 is not a point of inflection

d. $f\left(x\right)=x^3$

f”(x) = 6x

=> f”(0) = 0

This means x = 0 could be a point of inflection. To confirm this, we will find f’”(x) at x = 0

f”’(x) = 6

=> f’”(0) = 6

Since $f{”}^{\prime }\left(0\right)\ne 0$ and f”(0) =0, we can say x =0 is a point of inflection.

So the options a and d are correct