Prove there is real root in $[a;a-\frac{f(x)}{k}]$

Question

Let $f(x)$ be a continous function when $x \in [a;+\infty), f'(x)>k>0$ when $x\in (a;+\infty), k-\text{const}$. Prove there is only one real root in $(a;a-\frac{f(x)}{k})$ if $f(a)<0$.

My thoughts: as $f'(x)>0$ on $(a;+\infty)$ so $f(x)$ always increases. Hence, due to Intermediate Value Theorem, there is $c\in(a;a-\frac{f(x)}{k})$ such that $f(c)=0$ because $f(x)$ is a continuous function. So I need to prove that $f(a-\frac{f(x)}{k})>0$.

Answer 1

Using the fact that $f'(x)>k$ and Lagrange theorem we get $$f(a-\frac{f(x)}{k}) =f'(\xi)\cdot\frac{-f(a)}{k} + f(a) > k\cdot\frac{-f(a)}{k}+f(a)=0$$ $c$ is the only point because of the fact that $f'(x)>k>0$ $\forall x\in(a;+\infty)$