Science:Math Exam Resources/Courses/MATH100/December 2012/Question 03 (a)

MATH100 December 2012

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Question 03 (a)

Short-Answer Questions. Questions 1-4 are short-answer questions. Put your answers in the boxes provided. Simplify your answers as much as possible, and show your work. Each question is worth 3 marks, but not all questions are of equal difficulty. Suppose the tangent line to the curve $y=f(x)$ at $x=1$ passes through the points $(-2,3)$ and $(0,5)$. Find $\displaystyle f(1)$ and $\displaystyle f'(1)$.

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 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
First compute the equation of the tangent line. Then, how do the values of $f(1)$ and $f'(1)$ relate to this tangent line?

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We compute the equation of the line between the points $(-2,3)$ and $(0,5)$. First compute the slope via $m={\frac {5-3}{0-(-2)}}={\frac {2}{2}}=1.$ The y-intercept is given to be 5 since the point $(0,5)$ is on the line. Thus the equation of the line is $y=x+5$. As this line is tangent to the function at 1, we see that $\displaystyle m=1=f'(1)$ and $f(1)=y=(1)+5=6$ completing the question.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. According to the point slope formula given by $y-y_{0}=m(x-x_{0})$ we can use the fact that our tangent line in question contains the point $(1,f(1))$ and has slope $\displaystyle f'(1)$ and thus, we have the equation of the line is given by $\displaystyle y-f(1)=f'(1)(x-1)$ Now, we know that the points $(-2,3)$ and $(0,5)$ lie on our cruve and this gives the two equations ${\begin{aligned}5-f(1)=&f'(1)(0-1)\\3-f(1)=&f'(1)(-2-1)\end{aligned}}$ Simplifying gives ${\begin{aligned}5-f(1)=&-f'(1)\\3-f(1)=&-3f'(1)\end{aligned}}$ Subtracting yields ${\begin{aligned}2=&2f'(1)\end{aligned}}$ and so $\displaystyle f'(1)=1$. Plugging back into either of the original equations yields $f(1)=6$

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