Science:Math Exam Resources/Courses/MATH110/December 2016/Question 02 (a)

MATH110 December 2016

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Question 02 (a)
The tangent line to the graph of ${\textstyle y=f(x)}$ at the point where ${\textstyle x=4}$ has equation ${\textstyle 2x+3y=1}$. Find ${\textstyle f(4)}$ and ${\textstyle f'(4)}$.

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 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
Recall that the tangent line of a function $f$ at a point $x=a$ touches the graph of $f$ at the point $(a,f(a))$.

Hint 2
Recall that the derivative of a function $f$ at a point $x=a$ is the slope of its tangent line.

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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. By Hint 1, the tangent line at ${\textstyle x=4}$ touches the point ${\textstyle (4,f(4))}$. From the equation of the tangent line, since ${\textstyle f(4)}$ is the ${\textstyle y}$-value when ${\textstyle x=4}$, we solve $$2\cdot 4+3y=1$$ for ${\textstyle y}$ and obtain that $$8+3y=1\implies 3y=-7\implies y=-{\dfrac {7}{3}}.$$ Therefore, ${\textstyle f(4)=-{\dfrac {7}{3}}}$. To find ${\textstyle f'(4)}$, we just need the slope of the tangent line. Writing the equation of the tangent line as $$3y=-2x+1$$ $$y=-{\dfrac {2}{3}}x+{\dfrac {1}{3}},$$ the slope can be read off as ${\textstyle -{\dfrac {2}{3}}.}$ This means ${\textstyle f'(4)=-{\dfrac {2}{3}}}$. Answer: we have ${\textstyle f(4)={\color {blue}-{\dfrac {7}{3}}}}$ and ${\textstyle f'(4)={\color {blue}-{\dfrac {2}{3}}}}$.