Science:Math Exam Resources/Courses/MATH100/December 2015/Question 05 (b)

MATH100 December 2015

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Question 05 (b)
Let $f(x)$ be a function differentiable at $x=3$ and let $g(x)=x\cdot f(x)$. The line tangent to the curve $y=f(x)$ at $x=3$ has slope 2 while the line tangent to the curve $y=g(x)$ at $x=3$ has slope 5. What is $f(3)$?

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
For arbitrary given function $h(x)$, how to interpret the statement that the line tangent to the curve $y=h(x)$ at $x=a$ has slope $m$?

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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 line tangent to the curve $y=f(x)$ at $x=3$ has slope 2 means $f'(3)=2$. Similarly, since the line tangent to the curve $y=g(x)$ at $x=3$ has slope 5, we have $g'(3)=5$. Then, using the product rule with $g(x)=x\cdot f(x)$, $g'(x)=f(x)+x\cdot f'(x)$ , hence at $x=3$ we have $5=g'(3)=f(3)+3\cdot f'(3)=f(3)+6$. Therefore, the answer is $f(3)=-1$.