Science:Math Exam Resources/Courses/MATH104/December 2015/Question 01 (b)

MATH104 December 2015

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Question 01 (b)
Suppose $f(2)=3$ and $f'(2)=2.$ Let $h(x)={\frac {f(x)}{x^{2}}}.$ Find the equation of the line tangent to $y=h(x)$ at $x=2.$

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
We know that the tangent line must pass through $(2,h(2)),$ where it 'touches' the graph of $h(x).$ What is the slope of the line?

Hint 2
The slope of the tangent line we seek is $h'(2).$ However, we are not given an explicit expression for $h(x).$ How might we find $h'(2)$ using the information given about $f$ and its derivative?

Hint 3
Apply the quotient rule to find an expression for $h'(x)$ in terms of known (and/or computable) values.

Hint 4
Recall that the slope-point form of a line (the most convenient form here) is $y-y_{0}=m(x-x_{0}),$ where $m$ is the slope of the line and $(x_{0},y_{0})$ is a point the line passes through.

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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. Following the hints, we apply the quotient rule to $h(x)={\frac {f(x)}{x^{2}}}$ to find the slope of the tangent line $h'(2).$ ${\begin{aligned}h'(x)&={\frac {f'(x)\cdot x^{2}-f(x)\cdot 2x}{(x^{2})^{2}}}\\h'(2)&={\frac {f'(2)\cdot 2^{2}-f(2)\cdot 2(2)}{(2^{2})^{2}}}\\&={\frac {4f'(2)-4f(2)}{16}}\\&={\frac {4\cdot 2-4\cdot 3}{16}}\\&=-{\frac {1}{4}}.\end{aligned}}$ The tangent line passes through $(2,h(2))=(2,{\tfrac {f(2)}{2^{2}}})=({\color {OliveGreen}2},{\color {OrangeRed}{\tfrac {3}{4}}}),$ so the slope-point form of the equation of a line gives $y-{\color {OrangeRed}{\tfrac {3}{4}}}=-{\tfrac {1}{4}}(x-{\color {OliveGreen}2}).$ After rearranging, we obtain ${\color {blue}y=-{\tfrac {1}{4}}(x-2)+{\tfrac {3}{4}}}.$