Science:Math Exam Resources/Courses/MATH110/April 2012/Question 03 (b)

MATH110 April 2012

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Question 03 (b)
Consider a function ƒ whose derivative is equal to 0 on the entire interval $(-\infty ,\infty )$. Let x1 and x2 be two points on $(-\infty ,\infty )$. Prove, using your answer in part (a), that $\displaystyle f(x_{2})-f(x_{1})=0.$

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?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Does $f$ satisfy the conditions of Mean Value Theorem? What interval would you want to use to apply the Mean Value theorem?

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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. If $x_{1}=x_{2}$, the assumption is obviously true, so suppose $x_{1}\not =x_{2}$. Since $f'(x)=0$ everywhere, we can deduce that $f$ is differentiable everywhere, and hence also continuous everywhere, in particular on the interval $[x_{1},x_{2}]$. We can then apply the Mean Value Theorem, which tells us that there exists some c in the interval $(x_{1},x_{2})$ such that $f'(c)={\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}$. However, we know $f'(x)=0$ everywhere, so $f'(c)=0$ and we have $0={\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}$ and by cross multiplication, $\displaystyle 0=f(x_{2})-f(x_{1})$.

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