# Looking at tangent lines

• Apr 16th 2010, 06:40 AM
Kipster1203
Looking at tangent lines
Suppose that $\displaystyle I\subset \mathbb{R}$ is an open interval and that $\displaystyle f''(x) \geq 0$ for all $\displaystyle x\in I$. If $\displaystyle c \in I$ , show that the part of the graph of $\displaystyle f$ on $\displaystyle I$ is never below the tangent line to the graph at $\displaystyle (c,f(c))$ .
• Apr 16th 2010, 09:31 AM
Drexel28
Quote:

Originally Posted by Kipster1203
Suppose that $\displaystyle I\subset \mathbb{R}$ is an open interval and that $\displaystyle f''(x) \geq 0$ for all $\displaystyle x\in I$. If $\displaystyle c \in I$ , show that the part of the graph of $\displaystyle f$ on $\displaystyle I$ is never below the tangent line to the graph at $\displaystyle (c,f(c))$ .

So, let $\displaystyle \varphi(x)=f(x)-f'(c)(x-c)-f(c)$. Then, $\displaystyle \varphi(x)$ is the difference between $\displaystyle f$ and the tangent line at $\displaystyle x=c$. So, what we want to do is show that $\displaystyle \varphi'(x)\leqslant 0,\text{ }x\leqslant c$ and $\displaystyle \varphi'(x)\geqslant 0,\text{ }x\geqslant c$. But, $\displaystyle \varphi'(x)=f'(x)-f'(c)$ and $\displaystyle \varphi''(x)$ and so if $\displaystyle x\leqslant c$ then $\displaystyle \varphi'(x)\leqslant \varphi'(c)=0$ and if $\displaystyle x\geqslant c$ then $\displaystyle \varphi'(x)\geqslant \varphi'(c)=0$. From prior comment the conclusion follows.

NOTE! The above is purposefully full of holes! Fill them in!