Science:Math Exam Resources/Courses/MATH110/December 2015/Question 06 (d)

MATH110 December 2015

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Question 06 (d)
Let $f(x)$ be a function differentiable at $x=1$ and let $g(x)={\frac {f(x)}{x^{2}}}$. The line tangent to the curve $y=f(x)$ at $x=1$ has slope $3$ while the line tangent to the curve $y=g(x)$ at $x=1$ has slope $4$. What is $f(1)$?

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Translate the slope of the tangent line to each of the functions at a given point in terms of their derivative at that point.

Hint 2
Using the quotient rule, write $g'(1)$ in terms of $f(1)$ and $f'(1)$.

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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 $x=1$ has slope $3$ means that $f'(1)=3$, and the line tangent to the curve $y=g(x)$ at $x=1$ has slope $4$ means that $g'(1)=4$. These suggest that we should find the derivative of $g(x)$ using the qoutiet rule. ${\begin{aligned}g(x)={\frac {f(x)}{x^{2}}}\Rightarrow g'(x)={\frac {x^{2}f'(x)-2xf(x)}{x^{4}}}&\Rightarrow g'(1)={\frac {f'(1)-2f(1)}{1}}\\&\Rightarrow 4={3-2f(1)}\\&\Rightarrow 1=-2f(1)\end{aligned}}$ Therefore, we obtain $\color {blue}f(1)=-{\frac {1}{2}}$