Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 06 (b)/Solution 1

As we mentioned in the Hint 1, the graph of $f(x-p)$ is obtained by translating the graph of $f$ horizontally by $p$. Note that the shape of the graph doesn't change under the translation.

From part (a), we know that $f$ is strictly increasing on $\left(-\infty ,{\frac {1}{2}}\right)\cup \left({\frac {3}{2}},\infty \right)$, and strictly decreasing on $\left({\frac {1}{2}},{\frac {3}{2}}\right)$. Also, $f$ has its maximum value $2$ at $x={\frac {1}{2}}$ and $x=2$. Based on this, we can draw the following graph of $f$;

From the graph we can easily see that if we move the graph of $f$ horizontally to the left (i.e.,p<0), the maximum occurs at $x=2$, because of $f(x)>f(2)$ for $x>2$. Therefore, for $p<0$, the function $g_{p}(x)=f(x-p)$ has its maximum $g_{p}(2)=f(2-p)$, so that $M(p)=f(2-p)$.

On the other hand, for any $x<2$, we have $f(x)\leq f(2)=2$ and actually $f(x)=2$ only occurs at $x={\frac {1}{2}}$ in the range $(-\infty ,2)$. This implies that when we move the graph of $f$ horizontally to the right (i.e.,p>0), if ${\frac {1}{2}}+p$ is in $[0,2]$, the maximum of $g_{p}(x)=f(x-p)$ is $2$. Therefore, for $0<p\leq {\frac {3}{2}}$, we get $M(p)=2$. (by part (a), when $p=0$, we also have $M(p)=2$.

Furthermore, if we move the graph of $f$ horizontally to the right by $p>{\frac {3}{2}}$, (i.e, $p+{\frac {1}{2}}>2$), the graph is strictly increasing on $[0,2]$, so that the maximum occurs at $x=2$. As a result, $M(p)=f(2-p)$ for $p>{\frac {3}{2}}$.

To sum, we have $\color {blue}M(p)={\begin{cases}2&0\leq p\leq {\frac {3}{2}}\\f(2-p)&otherwise\end{cases}}$