Science:Math Exam Resources/Courses/MATH104/December 2014/Question 06/Solution 1

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 $(0,\infty )$.