Science:Math Exam Resources/Courses/MATH100/December 2016/Question 04 (a)

MATH100 December 2016

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Question 04 (a)
Consider the line $y=4x+2$. To which of the following functions is it tangent at $x=1$? $f(x)=x^{2}+3x+2$ $f(x)=2{\sqrt {x+3}}+2$ $f(x)=x^{3}+x+2$ $f(x)=x^{3}+2x^{2}+3x$ $f(x)=x^{3}+x^{2}-x$ None of these

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
A line is tangent to the function $y=f(x)$ at $x=a$, if first the tangency point lies on both the line and the function, and also the slope of the line is the derivative of the function at $x=a$. Therefore, a function is tangent to the given line if it goes through $(1,6)$ and has derivative $4$ at that point.

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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. As stated in the Hint, since the tangency point lies both on the tangent line and on the graph of the function, the value of the function at $x=1$ must be $f(1)=4(1)+2=6$. So, we check each option to see whether $(1,6)$ satisfies the function equation. Three of the given choices satisfy this condition: $f(x)=x^{2}+3x+2$ $f(x)=2{\sqrt {x+3}}+2$ $f(x)=x^{3}+2x^{2}+3x$ The next thing to check is whether the derivative of these functions at $x=1$ is equal to $4$: $f'(x)=2x+3\Rightarrow f'(1)=5$ $f'(x)={\frac {1}{\sqrt {x+3}}}\Rightarrow f'(1)={\frac {1}{2}}$ $f'(x)=3x^{2}+4x+3\Rightarrow f'(1)=10$ None of them satisfies $f'(1)=4$, so the answer is none of these.