[Math] How to prove that a function is above a its tangent line using the mean value theorem

calculus

Question:

Let $f(x)$ be a function, $f(x)$ is concave up and has continuous derivative for all $x > 0$. Use the mean value theorem to show that the graph of $y = f(x)$ lies above the tangent line at $x = 0$.

I do not know how to use MVT to prove this, if someone could help me solve this it would be greatly appreciated.

Best Answer

Maybe this one helps..

Lemma: If $f$ is concave up on $(a,b)$ and $\xi \in (a,b)$, then for all $x\in (a,b)$ it holds: $$f(x) \ge f(\xi) + (x-\xi)\cdot f'(\xi). \tag{1}$$ *Notice that $y = f(\xi) +(x-\xi)\cdot f'(\xi)$ is the equation of the tangent of $f$ at $x = \xi$.

Proof: Since $f$ is concave up on $(a,b)$, it holds that $f'$ is non - decreasing on $(a,b)$.

  • Let $x,\xi \in (a,b)$ with $x\lt \xi$. Applying the Mean Value Theorem (MVT) on $(x,\xi)$, we have that $\exists \xi_1 \in (x,\xi)$ such that: $$\begin{array}[t]{ll}f'(\xi_1) = \dfrac{f(\xi) - f(x) }{\xi -x}\le f'(\xi),&\quad \text{since $\xi_1 \lt \xi$}\\ f(\xi)- f(x) \le f'(\xi) \cdot (\xi - x),& \quad \text{since $x < \xi$} \\[2ex] f(x) \ge f'(\xi) \cdot (\xi - x) + f(\xi) \end{array} $$

  • Let $x,\xi \in (a,b)$ with $x \gt \xi$. Applying again the MVT, we take the inequality $(1)$.

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