Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (a)/Solution 1

The tangent line at the point $(x,f(x))$ is parallel to the line with the equation $y=2x$ exactly when $f'(x)=2$. So it is enough to prove that there is a value of $x$ such that $f'(x)=2$. We would like to use the Mean Value Theorem. Let $a=-1$ and $b=2$. Then $b-a=3$, $f(2)-f(-1)=5-(-1)=6$, and ${\frac {f(b)-f(a)}{b-a}}=6/3=2$. Because $f(x)$ is differentiable everywhere, it is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Thus, by the Mean Value Theorem, there exists an $x$ such that $a<x<b$ and $f'(x)={\frac {f(b)-f(a)}{b-a}}$. And as ${\frac {f(b)-f(a)}{b-a}}=2$, that means that there exists an $x$ such that $a<x<b$ and $f'(x)=2$. This finishes the proof.