Question

Find the absolute maximum and minimum of
$f(t) = 4 {t}^{3} - 5 {t}^{2} - 8t + 3 \: on \: \pmb{[} - 1,3\pmb{]}.$
​

Answer 1

$\Huge\text{ \bold{42\ and\ -\dfrac{191}{27}} }$

$\huge\text{\underline{\underline{Question}}}$

Find the absolute maximum and minimum of $f(t)=4t^{3}-5t^{2}-8t+3$ on $[-1,3]$.

$\huge\text{\underline{\underline{Topic}}}$

$\Large\text{[Differentiation]}$

If the differentiation of a function exists, the function increases or decreases according to the differentiation. So, we can determine the increase and decrease to find the maximum and minimum.

$\rightarrow\large\text{Formula for differentiation}$

For a polynomial term, it is known for $\text{ \boxed{(x^{n})'=nx^{n-1}} }$.

$\rightarrow\large\text{Absolute maximum/minimum}$

Absolute maximum and minimum are the maximum or minimum values in an interval. Like above, we can determine the differentiation to find these.

$\huge\text{\underline{\underline{Explanation}}}$

$\large\bold{[Step\ 1.\ Derivative]}$

$\text{ f(t)=4t^{3}-5t^{2}-8t+3\Longrightarrow\boxed{f'(t)=12t^{2}-10t-8} }$

$\large\bold{[Step\ 2.\ Investigation\ of\ variance]}$

$\bold{Given\ derivative}$

$f'(t)=2(6t^{2}-5t-4)=2(3t-4)(2t+1)$

$\bold{Increasing\ interval}$

$\text{ \cdots\longrightarrow\boxed{f'(t)>0 in t<-\dfrac{1}{2} or t>\dfrac{4}{3}} }$

$\bold{Critical\ points}$

$\text{ \cdots\longrightarrow\boxed{f'(t)=0 in t=-\dfrac{1}{2} or t=\dfrac{4}{3}} }$

$\bold{Decreasing\ interval}$

$\text{ \cdots\longrightarrow\boxed{f'(t)<0 in -\dfrac{1}{2}<t<\dfrac{4}{3}} }$

$\bold{Differentiation\ table}$

$\large\begin{tabular}{| c | c | c |}\cline{1-3} f'(t) & f(t) &Interval\\\cline{1-3} f'(t)>0 & \nearrow & (\infty,-\dfrac{1}{2}) \\\cline{1-3} f'(t)=0 & \rightarrow & t=-\dfrac{1}{2} \\\cline{1-3} f'(t)<0 & \searrow & (-\dfrac{1}{2},\dfrac{4}{3}) \\\cline{1-3} f'(t)=0 & \rightarrow & t=\dfrac{4}{3} \\\cline{1-3} f'(t)>0 & \nearrow & (\dfrac{4}{3},\infty) \\\cline{1-3}\end{tabular}$

$\large\bold{[Step\ 3.\ Evaluation]}$

As we are given that the function is defined in $[-1,3]$, we consider both the endpoints.

$\bold{Left\ endpoint}$

$\cdots\longrightarrow f(-1)=2$

$\bold{Critical\ points}$

$\cdots\longrightarrow f\left(-\dfrac{1}{2}\right)=\dfrac{21}{4}$

$\bold{Critical\ points}$

$\cdots\longrightarrow f\left(\dfrac{4}{3}\right)=-\dfrac{191}{27}$

$\bold{Right\ endpoint}$

$\cdots\longrightarrow f(3)=42$

$\large\bold{[Final\ answer]}$

$\cdots\longrightarrow\bold{The\ absolute\ maximum\ is\ 42.}$

$\text{ \cdots\longrightarrow\bold{The\ absolute\ minimum\ is\ -\dfrac{191}{27}.} }$