Find the absolute maximum and absolute minimum values of the function : $f\left(x\right)={x}^{4}-8{x}^{2}-10$ on interval = [-4,1]

luipieduq3
2021-12-04
Answered

Find the absolute maximum and absolute minimum values of the function : $f\left(x\right)={x}^{4}-8{x}^{2}-10$ on interval = [-4,1]

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Prioned

Answered 2021-12-05
Author has **11** answers

Find the absolute maximum and absolute minimum values of the function $f\left(x\right)={x}^{4}-8{x}^{2}-10$ on the interval [-4,1] as follows.

On the interval [-4,1], the critical points of the function$f\left(x\right)={x}^{4}-8{x}^{2}-10$ are x=-2 and x=0.

Now evaluate the function at the critical points x=-2 and x=0 and at the endpoints x=-4 and x=1 as shown below.

$f(-4)={(-4)}^{4}-8{(-4)}^{2}-10=256-128-10=118$

$f(-2)={(-2)}^{4}-8{(-2)}^{2}-10=16-32-10=-26$

$f\left(0\right)={\left(0\right)}^{4}-8{\left(0\right)}^{2}-10=0-0-10=-10$

$f\left(1\right)={(-1)}^{4}-8{(-1)}^{2}-10=1-8-10=-17$

Comparing the above values, it is clear that the absolute maximum and minimum of f on the interval [-4,1] are,

Absolute maximum : (x,f(x))=(-4, 118)

Absolute minimum : (x,f(x))=(-2, -26)

On the interval [-4,1], the critical points of the function

Now evaluate the function at the critical points x=-2 and x=0 and at the endpoints x=-4 and x=1 as shown below.

Comparing the above values, it is clear that the absolute maximum and minimum of f on the interval [-4,1] are,

Absolute maximum : (x,f(x))=(-4, 118)

Absolute minimum : (x,f(x))=(-2, -26)

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Exponential Growth and Decay

Exponential growth and decay problems follow the model given by the equation$A\left(t\right)=P{e}^{rt}$

-The model is a function of time t

-A(t) is the amount we have ater time t

-PIs the initial amount, because for t=0, notice how$A\left(0\right)=P{e}^{0\times t}=P{e}^{0}=P$

-Tis the growth or decay rate. It is positive for growth and negative for decay

Growth and decay problems can deal with money (interest compounded continuously), bacteria growth, radioactive decay. population growth etc.

So A(t) can represent any of these depending on the problem.

Practice

The growth of a certain bactenia population can be modeled by the function

$A\left(t\right)=900{e}^{0.0534}$

where A(t) is the number of bacteria and t represents the time in minutes.

How long will t take for the number of bacteria to double? (your answer must be accurate to at least 3 decimal places.)

Exponential growth and decay problems follow the model given by the equation

-The model is a function of time t

-A(t) is the amount we have ater time t

-PIs the initial amount, because for t=0, notice how

-Tis the growth or decay rate. It is positive for growth and negative for decay

Growth and decay problems can deal with money (interest compounded continuously), bacteria growth, radioactive decay. population growth etc.

So A(t) can represent any of these depending on the problem.

Practice

The growth of a certain bactenia population can be modeled by the function

where A(t) is the number of bacteria and t represents the time in minutes.

How long will t take for the number of bacteria to double? (your answer must be accurate to at least 3 decimal places.)

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$2x-3y+4z=-19\phantom{\rule{0ex}{0ex}}6x+4y-2z=8\phantom{\rule{0ex}{0ex}}x+5y+4z=23$

what I have done so far is I put the nubmer and x,y and z in matrix form:

$\left[\begin{array}{ccc}2& -3& 4\\ 6& 4& -2\\ 1& 5& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-19\\ 8\\ 23\end{array}\right]$

what I have done so far is I put the nubmer and x,y and z in matrix form:

$\left[\begin{array}{ccc}2& -3& 4\\ 6& 4& -2\\ 1& 5& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-19\\ 8\\ 23\end{array}\right]$

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$\{\begin{array}{l}-x+2z=0\\ 3x-6z=0\\ 2x-4z=0\end{array}$

First of all, I don't see any $y$ variable there. I suppose it doesn't matter and I proceed normally:

$\left[\begin{array}{ccc}-1& 2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-{f}_{1}$

$\left[\begin{array}{ccc}1& -2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-3{f}_{1}+{f}_{2}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 2& -4& 0\end{array}\right]$

$-2{f}_{1}+{f}_{3}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

So, this is the staggered reduced form.

This is an homogeneous system (because of the null column), thus one solution is $(0,0,0)$.

Other than that, I have to check out the range of the system. The range is $1$, which is less than the number of columns... what is the number of columns?

$\{\begin{array}{l}-x+2z=0\\ 3x-6z=0\\ 2x-4z=0\end{array}$

First of all, I don't see any $y$ variable there. I suppose it doesn't matter and I proceed normally:

$\left[\begin{array}{ccc}-1& 2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-{f}_{1}$

$\left[\begin{array}{ccc}1& -2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-3{f}_{1}+{f}_{2}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 2& -4& 0\end{array}\right]$

$-2{f}_{1}+{f}_{3}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

So, this is the staggered reduced form.

This is an homogeneous system (because of the null column), thus one solution is $(0,0,0)$.

Other than that, I have to check out the range of the system. The range is $1$, which is less than the number of columns... what is the number of columns?

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Suppose ${m}_{x},{M}_{x},{m}_{y},{M}_{y}\in \mathbb{R}$ are known and ${x}_{l},{x}_{h},{y}_{l},{y}_{h}\in \mathbb{R}$ are unknown. Does the simultaneous system

$\begin{array}{rl}{x}_{l}\phantom{\rule{1mm}{0ex}}+\phantom{\rule{1mm}{0ex}}{x}_{h}& \phantom{\rule{1mm}{0ex}}=\phantom{\rule{1mm}{0ex}}{y}_{l}+\phantom{\rule{1mm}{0ex}}{y}_{h}\\ {m}_{x}\phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{x}_{l}& \phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{x}_{h}\phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{M}_{x}\\ {m}_{y}\phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{y}_{l}& \phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{y}_{h}\phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{M}_{y}\end{array}$

have a unique solution $({x}_{l},{x}_{h},{y}_{l},{y}_{h})$

$\begin{array}{rl}{x}_{l}\phantom{\rule{1mm}{0ex}}+\phantom{\rule{1mm}{0ex}}{x}_{h}& \phantom{\rule{1mm}{0ex}}=\phantom{\rule{1mm}{0ex}}{y}_{l}+\phantom{\rule{1mm}{0ex}}{y}_{h}\\ {m}_{x}\phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{x}_{l}& \phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{x}_{h}\phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{M}_{x}\\ {m}_{y}\phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{y}_{l}& \phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{y}_{h}\phantom{\rule{1mm}{0ex}}\le \phantom{\rule{1mm}{0ex}}{M}_{y}\end{array}$

have a unique solution $({x}_{l},{x}_{h},{y}_{l},{y}_{h})$