Ques. -29

Calculus Level 2

Let $f$ be a twice differentiable function satisfying $f(1)=1, f(2)=4, f(3)=9$ , then :

f''(x)=2 \forall x \in \mathbb{R}

There exists at least one

x \in (1,3)

such that

f''(x)=2

. None of the given is true.

f'(x)=5 \forall x \in \mathbb{R}

2 solutions

Samer Atasi
Apr 22, 2016

The function g(x) = f(x) - x^2 is equal to 0 for x= 1, x=2 and x=3. This implies that g(x) has 2 extremas, one in each interval (1,2) and (2,3). Between these 2 extremas the second derivative g''(x) must change sign and therefore pass through zero at some point X. At this point X, g''(X)=0, therefore f''(X)=2.

According to the question, f is a function, not specifically told that it is a polynomial, so none of the options is correct.

A Former Brilliant Member - 4 years, 10 months ago

I agree with you. Even if it is a polynomial, it could be of the third or more degree (those still satisfy "twice-differentiable").

Exponent Bot - 3 years ago

Edsel Adap
Dec 3, 2015

Given f(1) = 1, f(2) = 4, f(3)=9; it is plausible that f(x)=x^2 Suppose f(x) = x^2 then f'(x)=2x and f''(x)=2.

choose an x from the set (1,3), and it can be shown to satisfy both the f(x) and the f''(x) we came up with.

why should we chose value of x from 1 to 3 when the function is defined for all real numbers?

Rudransh Srivastava - 5 years, 6 months ago

Although it is plausible that f (x) = x^2, assuming that it is so seems to me not right. It could be some other function. I thought f (x) = x^2, but I like doing Mathematics under as little(or no) assumptions as possible.

Mita Ramabulana - 5 years, 6 months ago

According to the question, f is a function, not specifically told that it is a polynomial, so none of the options is correct.

A Former Brilliant Member - 4 years, 10 months ago

Ques. -29

2 solutions

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