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

MATH104 December 2014

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Question 01 (d)
Let $f(x)={\frac {x^{3}}{3}}+x^{2}-7x+100$. Find the point on the graph of $f$ at which the tangent line has slope -8.

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 $f$. What does it represent?

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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. Taking the derivative yields $f'(x)=x^{2}+2x-7.$ The derivative of $f$ at $x$ is the slope of the tangent line at $(x,f(x))$. So we set $f'(x)=-8$ and solve the quadratic equation using the quadratic formula/factoring: $\displaystyle {\begin{aligned}x^{2}+2x-7&=-8\\x^{2}+2x+1&=0\\(x+1)^{2}&=0\end{aligned}}$ So the $f$ has tangent line of slope $-8$ when $x=-1$. The point which we get is $(-1,f(-1))=\left(-1,{\frac {107\cdot 3+2}{3}}\right)=\left(-1,{\frac {323}{3}}\right)$, where $f(-1)={\frac {(-1)^{3}}{3}}+(-1)^{2}-7(-1)+100={\frac {107\cdot 3+2}{3}}={\frac {323}{3}}$.

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