If $f(x)$ is a twice differentiable function and given that $f(1) = 2, f(2) = 5$ and $f(3) = 10$ then

Question

Q. If $f(x)$ is a twice differentiable function and given that $f(1) = 2, f(2) = 5$ and $f(3) = 10$ then

(A) $f''(x) = 2$ for all $x\in (1,3)$

(B) $f''(x) = f'(x) = 2$ for some $x\in (2,3)$

(C) $f''(x) = 3$ for all $x\in (2,3)$

(D) $f''(x) = 2$ for some $x\in (1,3)$

I assumed $$f(x) = ax^2 + bx + c$$ and compared and found $a,b,c$. That way, I get $f''(x) = 2$ for all $x$. But, seems like that's not how it is done. I am aware that it could be a thrce differentiable function or a trigonometric one. But, I thought that a general case should satisfy it and hence I could atleast deduce the answer. Why is it wrong? What should I do?

Answer 1

Hint:

If $ f$ is a continuous function on the closed interval $ [a,b]$, and differentiable on the open interval $(a,b)$, then there exists a point $c $ in $ (a,b)$ such that

$$ f'(c)=\frac {f(b)-f(a)}{b-a}$$

Answer 2

The mistake you have made is: Assuming that the function is a polynomial function. Polynomial function is just a special case.

There could certainly be another strange function like a trigonometric function satisfying the given conditions.

Hence the answer should be D and not B

Answer 3

If $f''$ is a constant for all $x$ within an interval, then $f$ is a quadratic. Following your approach you would get that $f(x)=x^2+1$. The second derivative of this is $2$ which rules $(C)$ out.

However, there are functions that are more than just polynomials that satisfy the three conditions. In fact, there is a twice-differentiable function that does so of the form $f(x)=Ax^2+Bx+C\sin x+D\cos x$. It can clearly be seen that $f''(x)\ne 2$ for all $x\in(1,3)$. Thus rules $(A)$ out.

Finally, if you use that function, it is possible to show that between $(2,3)$, $f'(x)\ne f''(x)$ at any point*. This rules $(B)$ out.

*: since $f'(2)<f''(2)$ and $f'(3)<f''(3)$, if suffices to prove that between $(2,3)$, $f'$ is monotonically increasing.

$$\textsf{The answer is (D).}$$

Answer 4

Only (D) can possibly be correct, since $f(x) = 2^x + x-1$ violates (A), (B), (C).

Added: (D) must be correct. To see this, write any such $f$ as $$ f(x) = x^2 + 1 + g(x) \, . $$ Then $g(1) = g(2) = g(3) = 0$. Therefore $g$ cannot be convex or concave on $[1,3]$ and there must be a $c \in (1,3)$ where $g''(c) = 0$. Thus $f''(c) = 2$.