# If $f : R\rightarrow R$ is a twice differentiable function such that $f''(x) > 0$ $\forall$ $x\in R$ then which of the following is correct?

If $$f : R\rightarrow R$$ is a twice differentiable function such that $$f''(x) > 0$$ $$\forall$$ $$x\in R$$ and $$f\left(\dfrac{1}{2}\right)=\dfrac{1}{2},f(1)=1$$, then which of the following is correct:-

i) $$f'(1)\le0$$

ii) $$\dfrac{1}{2}< f'(1)\le1$$

iii) $$f'(1)>1$$

iv) $$0

My attempt is as follows:-

$$f''(x)>0$$

Integrating both sides

$$f'(x)>c_1$$

Again integrating on both sides

$$f(x)>c_1x+c_2$$

$$f\left(\dfrac{1}{2}\right)>\dfrac{c_1}{2}+c_2$$ $$1>c_1+2c_2\tag{1}$$

$$f(1)>c_1+c_2$$ $$1>c_1+c_2\tag{2}$$

Subtracting $$(2)$$ from $$(1)$$

$$c_2<0\tag{3}$$

Multiplying $$(2)$$ with $$2$$ and then subtracting from $$(1)$$

$$-1>-c_1$$ $$c_1>1\tag{4}$$

As we know $$f'(x)>c_1$$, so we can say $$f'(x)>1$$.

So answer should be $$f'(1)>1$$. But is this correct way of solving this question? Is this the correct method or am I violating something here? I doubt it because this question has been solved using Lagrange's theorem. Please help me in this.

• It is not true that $f''(x)>0$ implies $f'(x)>c$ for all $x$. However, you can integrate from $x_0$ to $x_1$ and conclude that $f'(x_1)-f'(x_0)>0$ for $x_1>x_0$. – almagest Dec 17 '19 at 10:38

Your argument is not correct. For example the second derivative of $$x^{2}$$ is greater than $$0$$ but you cannot say $$2x>c_1$$ for all real numbers $$x$$.
By MVT we have $$1=\frac {f(1)-f(\frac 1 2)} {1-\frac 1 2}=f'(t)$$ for some $$t \in (\frac 1 2, 1)$$. Since $$f'' >0$$ implies that $$f'$$ is strictly increasing we get $$1 =f'(t) . So iii) is true and all other options are false.