Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 06 (a)

MATH102 December 2016

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Question B 06 (a)
For both parts of this question, consider the function $f(x)=4x^{3}-12x^{2}+9x.$ (a) Find the absolute maximum of the function $f(x)$ on the interval $[0,2]$. Hint: It would be useful to make a rough sketch of $f(x)$

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Hint
Examine the signs of $f'$ on $[0,2]$.

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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. To see on which interval $f$ is increasing ore decreasing, we first determine the sign of $f'$ on $[0,2]$. For this purpose, we find the derivative $f'(x)=(4x^{3}-12x^{2}+9x)'=12x^{2}-24x+9=3(4x^{2}-8x+3)=3(4(x^{2}-2x+1)-1)=3(4(x-1)^{2}-1)$. Since $4(x-1)^{2}-1=0\iff (x-1)^{2}={\frac {1}{4}}\iff x=1\pm {\frac {1}{2}}\iff x={\frac {1}{2}}{\text{ and }}x={\frac {3}{2}}$, drawing the graph of $f'(x)$, we have $f'(x)>0$ when $x<{\frac {1}{2}}$ or $x>{\frac {3}{2}}$, and $f'(x)<0$ when ${\frac {1}{2}}<x<{\frac {3}{2}}$. Therefore, $f$ is increasing on $(-\infty ,{\frac {1}{2}})\cup ({\frac {3}{2}},\infty )$ and decreasing on $({\frac {1}{2}},{\frac {3}{2}})$. This means a possible points to achieve absolute maximum of $f$ on $[0,2]$ are the critical point $x={\frac {1}{2}}$ and the end point $x=2$. Comparing $f\left({\frac {1}{2}}\right)=4\cdot \left({\frac {1}{2}}\right)^{3}-12\cdot \left({\frac {1}{2}}\right)^{2}+9\cdot \left({\frac {1}{2}}\right)=2$ with $f(2)=(4\cdot 2^{3}-12\cdot 2^{2}+9\cdot 2)=2$, we can figure out that $f$ achieve its maximum on $[0,2]$ at $x={\frac {1}{2}},2$, which is $\color {blue}f\left({\frac {1}{2}}\right)=f(2)=2$.