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

MATH110 April 2016

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Question 09 (a)
Suppose that the function $f$ is differentiable for all $x$ and that $f(-1)=-1$ and $f(2)=5$. (a) Prove that there is a point on the graph of $f$ at which the tangent line is parallel to the line with the equation $y=2x$. Make sure you state what theorem(s) you use to support your claim and why they apply.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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Hint
Recall the Mean Value Theorem, and explain how this theorem can be applied to this question.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.