Science:Math Exam Resources/Courses/MATH104/December 2014/Question 06

MATH104 December 2014

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Question 06
Let $f(x)={\frac {x^{4}}{12}}-{\frac {4(3^{-x})}{(\ln {3})^{2}}}+\log _{(x+1)^{2}}(x+1)^{2014}$ be a function defined on the interval $(0,\infty )$. Show that the graph of $f(x)$ on the interval $(0,\infty )$ has an inflection point.

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
First simplify the function by using the change the base formula for logarithms.

Hint 2
Think about what the necessary and sufficient conditions are for there to be an inflection point.

Hint 3
Additionally, Intermediate Value Theorem will be very very useful here.

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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. The necessary condition for $f$ to have an inflection point is that there exists a $c$ in $(0,\infty )$ so that $f''(c)=0$. A sufficient condition for $f$ to have an inflection point is that $f''$ changes signs at $c$. First, let us simplify the function $\log _{(x+1)^{2}}(x+1)^{2014}$. Using the change of base of logarithms formula given by $\log _{d}(a)=\log _{d}(b)\log _{b}(a)$. Take $d=e$, $b=(x+1)^{2}$, and $a=(x+1)^{2014}$, we obtain that $\log _{(x+1)^{2}}(x+1)^{2014}={\frac {\ln {(x+1)^{2014}}}{\ln {(x+1)^{2}}}}={\frac {2014\ln {(x+1)}}{2\ln(x+1)}}=1007$ Thus $f(x)={\frac {x^{4}}{12}}-{\frac {4(3^{-x})}{(\ln {3})^{2}}}+1007$. Now we take the first and second derivatives of $f$. Recall that if $g(x)=3^{-x}$, then $g'(x)=-3^{-x}\ln {3}$. Taking the first derivative of $f$, we get: $f'(x)={\frac {4x^{3}}{12}}+{\frac {4(3^{-x})\ln {3}}{(\ln {3})^{2}}}={\frac {x^{3}}{3}}+{\frac {4(3^{-x})}{(\ln {3})}}$. Now, we take the second derivative of $f$ and find: $f''(x)=x^{2}-{\frac {4(3^{-x})\ln {3}}{(\ln {3})}}=x^{2}-4(3^{-x})$. Remark next that $f''(x)$ is a continuous function of $x$ on the interval $(0,\infty )$, since both $x^{2}$ and $3^{-x}$ are continuous functions of $x$ on $(0,\infty )$. Now we want to show that there exists a $c\in (0,\infty )$ so that $f''(c)=0$. Evaluate $f''(0)$ gives: $f''(0)=0^{2}-4=-4<0$. Now evaluate $f''(2)$ to get: $f''(2)=(2)^{2}-4(3^{-2})=2^{2}-{\frac {4}{9}}>0$. Since $f''(x)$ is continuous on $(0,\infty )$, and $f''(0)<0$ while $f''(2)>0$, by the Intermediate Value Theorem, there exists a point $c\in (0,2)$ so that $f''(c)=0$. We have already shown that there is a point to the left and to the right of c with different signs. So indeed the given function $f$ has an inflection point on $(0,\infty )$.

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