Science:Math Exam Resources/Courses/MATH104/December 2014/Question 01 (e)

MATH104 December 2014

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Question 01 (e)
Suppose $f(4)=3$ and $f'(4)=2$. Let $h(x)={\frac {f(x)}{x-5}}$. Find an equation of the line tangent to $y=h(x)$ at $x=4$.

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
Take the derivative of $h$ using quotient rule.

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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. Note first that $h(4)={\frac {f(4)}{4-5}}=-3$. Rewrite $h(x)=f(x)(x-5)^{-1}$ and take the derivative of $h$ with respect to $x$ using the product rule: $h'(x)=f'(x)(x-5)^{-1}+f(x)(-1)(x-5)^{-2}$ Alternativele, if you’re more comfortable with the quotient rule: ${\begin{aligned}h'(x)&={\frac {f'(x)(x-5)-f(x)(1)}{(x-5)^{2}}}\\&={\frac {f'(x)}{(x-5)}}-{\frac {f(x)}{(x-5)^{2}}}\end{aligned}}$ Evaluating $h'$ at $x=4$ gives the slope of the tangent line at $x=4$ to be $h'(4)=f'(4)(4-5)^{-1}+f(4)(-1)(4-5)^{-2}=-2-3=-5$ We know that $(4,h(4))$ is a point on the tangent line of $h(x)$ at $x=4$; so we obtain $y-h(4)=-5(x-4)$ which gives $y=-5x+20+h(4)=-5x+20+(-3)=-5x+17$

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