Maxima and Minima

Question 1.

Find the maximum and minimum values, if any, of the following functions:

(i) $f(x)=(2x\text{–}1{)}^2+3$
(ii) $f(x)=9x^2+12x+2$
(iii) $f(x)=\text{–}(x\text{–}1{)}^2+10$
(iv) $g(x)=x^3+1$

Solutions

(i) $f(x)=(2x\text{–}1{)}^2+3$

$$f(x)=(2x-1{)}^2+3$$

This is a quadratic in the form $a(x-h{)}^2+k$ where $a=4>0$, hence it opens upwards, so it has a minimum.

Minimum occurs when the squared term is zero:

$$(2x-1{)}^2=0\Rightarrow x=\frac{1}{2}$$

$$f\left(\frac{1}{2}\right)=3$$

Minimum value = 3 at $x=\frac{1}{2}$
No maximum

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(ii) $f(x)=9x^2+12x+2$

$$f(x)=9x^2+12x+2$$

Use $x=-\frac{b}{2a}$

$$x=-\frac{12}{2\cdot 9}=-\frac{2}{3}$$

Now substitute:

$$f(-\frac{2}{3})=9\left(\frac{4}{9}\right)+12(-\frac{2}{3})+2=4-8+2=-2$$

Minimum value = –2 at $x=-\frac{2}{3}$
No maximum

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(iii) $f(x)=\text{–}(x\text{–}1{)}^2+10$

This is a downward opening parabola ($a=-1<0$), so it has a maximum.

Maximum occurs when squared term is zero:

$$x-1=0\Rightarrow x=1$$

$$f(1)=10$$

Maximum value = 10 at $x=1$
No minimum

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(iv) $g(x)=x^3+1$

Find derivative:

$$g^{\mathrm{\prime }}(x)=3x^2$$

$$g^{\mathrm{\prime }}(x)=0\Rightarrow x=0$$

Second derivative:

$$g^{\mathrm{\prime }\mathrm{\prime }}(x)=6x,g^{\mathrm{\prime }\mathrm{\prime }}(0)=0$$

This is a point of inflection, not a maximum/minimum.

So:

No maximum or minimum values (cubic increases from $-\mathrm{\infty }$ to $+\mathrm{\infty }$).

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Question 2.

Find the maximum and minimum values, if any, of the following functions:

(i) $f(x)=\mathrm{\mid }x+2\mathrm{\mid }-1$
(ii) $g(x)=-\mathrm{\mid }x+1\mathrm{\mid }+3$
(iii) $h(x)=\sin (2x)+5$
(iv) $f(x)=\mathrm{\mid }\sin 4x+3\mathrm{\mid }$
(v) $h(x)=x+1,x\in (-1,1)$

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Solutions

(i) $f(x)=\mathrm{\mid }x+2\mathrm{\mid }-1$

The function $\mathrm{\mid }x+2\mathrm{\mid }$has a minimum value 0 at $x=-2$.

So,

$$f(-2)=0-1=-1$$

Minimum value = –1 at $x=-2$
No maximum (because $\mathrm{\mid }x+2\mathrm{\mid }\to \mathrm{\infty }\Rightarrow f(x)\to \mathrm{\infty }$)

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(ii) $g(x)=-\mathrm{\mid }x+1\mathrm{\mid }+3$

$\mathrm{\mid }x+1\mathrm{\mid }$ has minimum value 0 at $x=-1$.
So,

$$g(-1)=-(0)+3=3$$

Maximum value = 3 at $x=-1$
No minimum (since $-\mathrm{\mid }x+1\mathrm{\mid }\to -\mathrm{\infty }$)

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(iii) $h(x)=\sin (2x)+5$

We know that:

$$-1\le \sin (2x)\le 1$$

Add 5 to each part:

$$4\le \sin (2x)+5\le 6$$

Thus:

Minimum value = 4
Maximum value = 6

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(iv) $f(x)=\mathrm{\mid }\sin 4x+3\mathrm{\mid }$

$$-1\le \sin 4x\le 1$$

Add 3:

$$2\le \sin 4x+3\le 4$$

Since absolute value is applied:

$$\mathrm{\mid }\sin 4x+3\mathrm{\mid }=\sin 4x+3(\text{always positive already})$$

Therefore:

Minimum value = 2
Maximum value = 4

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(v) $h(x)=x+1,x\in (-1,1)$

This is a linear function.

Evaluate at interval boundaries:

Left end: $x\to -1$,

$$h(x)\to (-1+1)=0(\text{not included})$$

Right end: $x\to 1$,

$$h(x)\to (1+1)=2(\text{not included})$$

Since interval is open, values 0 and 2 are not attained.

No maximum and no minimum
(Values approach 0 and 2 but never reach them)

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Question 3.

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i) $f(x)=x^2$
(ii) $g(x)=x^3\text{–}3x$
(iii) $h(x)=\sin x+\cos x,0<x<\frac{\pi }{2}$
(iv) $f(x)=\sin x-\cos x,0<x<2\pi$
(v) $f(x)=x^3-6x^2+9x+15$
(vi) $g(x)=\frac{x}{2}+\frac{2}{x},x>0$
(vii) $g(x)=\frac{1}{x^2+2}$
(viii) $f(x)=x\sqrt{1-x},0<x<1$

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Solutions

(i) $f(x)=x^2$

$$f^{\mathrm{\prime }}(x)=2x=0\Rightarrow x=0$$

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=2>0$$

So local minimum at $x=0$

$$f(0)=0$$

Local minimum value = 0 at $x=0$
No local maximum.

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(ii) $g(x)=x^3\text{–}3x$

$$g^{\mathrm{\prime }}(x)=3x^2-3=3(x^2-1)=0\Rightarrow x=\pm 1$$

$$g^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$$

At $x=-1$, $g^{\mathrm{\prime }\mathrm{\prime }}(-1)=-6<0$ → local maximum

$$g(-1)=(-1{)}^3-3(-1)=2$$

At $x=1$, $g^{\mathrm{\prime }\mathrm{\prime }}(1)=6>0$ → local minimum

$$g(1)=1-3=-2$$

Local maximum value = 2 at $x=-1$
Local minimum value = –2 at $x=1$

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(iii) $h(x)=\sin x+\cos x,0<x<\frac{\pi }{2}$

$$h^{\mathrm{\prime }}(x)=\cos x-\sin x=0\Rightarrow \cos x=\sin x\Rightarrow x=\frac{\pi }{4}$$

$$h^{\mathrm{\prime }\mathrm{\prime }}(x)=-\sin x-\cos x$$

At $x=\frac{\pi }{4}$,

$$h^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{\pi }{4}\right)=-\sqrt{2}<0$$

So local maximum.$$h\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$$

Local maximum value = $\sqrt{2}$ at $x=\frac{\pi }{4}$
No local minimum in given interval.

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(iv) $f(x)=\sin x-\cos x,0<x<2\pi$

$$f^{\mathrm{\prime }}(x)=\cos x+\sin x=0\Rightarrow \tan x=-1$$

Solutions in interval:

$$x=\frac{3\pi }{4},\frac{7\pi }{4}$$

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=-\sin x+\cos x$$

At $x=\frac{3\pi }{4}$,

$$f^{\mathrm{\prime }\mathrm{\prime }}<0\Rightarrow \text{local maximum}$$

$$f\left(\frac{3\pi }{4}\right)=\frac{\sqrt{2}}{2}-(-\frac{\sqrt{2}}{2})=\sqrt{2}$$

At $x=\frac{7\pi }{4}$,

$$f^{\mathrm{\prime }\mathrm{\prime }}>0\Rightarrow \text{local minimum}$$
$$f\left(\frac{7\pi }{4}\right)=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}$$

Local maximum = $\sqrt{2}$ at $x=\frac{3\pi }{4}$
Local minimum = $-\sqrt{2}$ at $x=\frac{7\pi }{4}$

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(v) $f(x)=x^3-6x^2+9x+15$

$$f^{\mathrm{\prime }}(x)=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3)=0\Rightarrow x=1,3$$

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x-12$$

At $x=1$, $f^{\mathrm{\prime }\mathrm{\prime }}(1)=-6<0$ → local maximum

$$f(1)=1-6+9+15=19$$

At $x=3$, $f^{\mathrm{\prime }\mathrm{\prime }}(3)=6>0$ → local minimum

$$f(3)=27-54+27+15=15$$

Local maximum value = 19 at $x=1$
Local minimum value = 15 at $x=3$

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(vi) $g(x)=\frac{x}{2}+\frac{2}{x},x>0$

$$g^{\mathrm{\prime }}(x)=\frac{1}{2}-\frac{2}{x^2}=0\Rightarrow \frac{2}{x^2}=\frac{1}{2}\Rightarrow x=2$$

$$g^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{4}{x^3}\Rightarrow g^{\mathrm{\prime }\mathrm{\prime }}(2)=\frac{4}{8}>0$$

So local minimum at $x=2$

$$g(2)=1+1=2$$

Local minimum value = 2 at $x=2$
No local maximum.

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(vii) $g(x)=\frac{1}{x^2+2}$

$$g^{\mathrm{\prime }}(x)=-\frac{2x}{(x^2+2{)}^2}=0\Rightarrow x=0$$

$$g^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{6x^2-4}{(x^2+2{)}^3}\Rightarrow g^{\mathrm{\prime }\mathrm{\prime }}(0)=-\frac{4}{8}<0$$

So local maximum at $x=0$$$g(0)=\frac{1}{2}$$

Local maximum value = $\frac{1}{2}$ at $x=0$
No minimum.

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(viii) $f(x)=x\sqrt{1-x},0<x<1$

$$f^{\mathrm{\prime }}(x)=\sqrt{1-x}+x(-\frac{1}{2\sqrt{1-x}})=\frac{2(1-x)-x}{2\sqrt{1-x}}$$

Set numerator zero:

$$2-2x-x=0\Rightarrow 2-3x=0\Rightarrow x=\frac{2}{3}$$
$$f\left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3}\sqrt{\frac{1}{3}}=\frac{2}{3\sqrt{3}}$$

$$f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{2}{3}\right)<0\Rightarrow \text{local maximum}$$

Local maximum value = $\frac{2}{3\sqrt{3}}$ at $x=\frac{2}{3}$
No minimum in interval.

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Question 4.

Prove the following functions do not have maxima or minima:

(i) $f(x)=e^x$
(ii) $g(x)=\log x$
(iii) $h(x)=x^3+x^2+x+1$

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Solutions

(i) $f(x)=e^x$

$$f(x)=e^x$$

Differentiate:

$$f^{\mathrm{\prime }}(x)=e^x>0\text{for all real }x$$

Since derivative is always positive, function is strictly increasing on $(-\mathrm{\infty },\mathrm{\infty })$.

Therefore, it never turns back to form a peak (maximum) or valley (minimum).

Hence, $e^x$ has no maximum and no minimum.

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(ii) $g(x)=\log x$, $x>0$

$$g^{\mathrm{\prime }}(x)=\frac{1}{x}>0\text{for all }x>0$$

Derivative is always positive in its domain, so $\log x$ is strictly increasing.

Therefore, $\log x$ has no maxima or minima.

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(iii) $h(x)=x^3+x^2+x+1$

$$h(x)=x^3+x^2+x+1$$

Differentiate:$$h^{\mathrm{\prime }}(x)=3x^2+2x+1$$

Check discriminant of the quadratic:

$$\mathrm{\Delta }=(2{)}^2-4(3)(1)=4-12=-8<0$$

Since discriminant < 0 → quadratic has no real roots, and the coefficient of $x^2$ is positive,

$$3x^2+2x+1>0\text{ for all }x$$

Thus derivative is always positive, so the function is strictly increasing.

Therefore, $h(x)$ has no maxima or minima.

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Question 5.

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) $f(x)=x^3,x\in [-2,2]$
(ii) $f(x)=\sin x+\cos x,x\in [0,\pi ]$
(iii) $f(x)=4x-\frac{1}{2}{x}^2,x\in [-2,\frac{9}{2}]$
(iv) $f(x)=(x-1{)}^2+3,x\in [-3,1]$

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Solutions

(i) $f(x)=x^3$, $x\in [-2,2]$

Derivative:

$$f^{\mathrm{\prime }}(x)=3x^2=0\Rightarrow x=0$$

Check values at critical point and endpoints:

$$f(-2)=(-2{)}^3=-8$$

$$f(0)=0$$

$$f(2)=8$$

Answer:

• Absolute maximum = 8 at $x=2$

• Absolute minimum = –8 at $x=-2$

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(ii) $f(x)=\sin x+\cos x,x\in [0,\pi ]$

Derivative:

$$f^{\mathrm{\prime }}(x)=\cos x-\sin x=0\Rightarrow \sin x=\cos x\Rightarrow x=\frac{\pi }{4}$$

Evaluate:

$$f\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$$

Check endpoints:

$$f(0)=\sin 0+\cos 0=1$$

$$f(\pi )=\sin \pi +\cos \pi =-1$$

Answer:

• Absolute maximum = $\sqrt{2}$ at $x=\frac{\pi }{4}$

• Absolute minimum = –1 at $x=\pi$
  

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(iii) $f(x)=4x-\frac{1}{2}{x}^2,x\in [-2,\frac{9}{2}]$

Derivative:

$$f^{\mathrm{\prime }}(x)=4-x=0\Rightarrow x=4$$

Evaluate at critical point and endpoints:

$$f(-2)=4(-2)-\frac{1}{2}(4)=-8-2=-10$$

$$f(4)=4(4)-\frac{1}{2}(16)=16-8=8$$

$$f\left(\frac{9}{2}\right)=4\cdot \frac{9}{2}-\frac{1}{2}\left(\frac{81}{4}\right)=18-\frac{81}{8}=18-10.125=7.875$$

Answer:

• Absolute maximum = 8 at $x=4$

• Absolute minimum = –10 at $x=-2$

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(iv) $f(x)=(x-1{)}^2+3,x\in [-3,1]$

Derivative:

$$f^{\mathrm{\prime }}(x)=2(x-1)=0\Rightarrow x=1$$

Check endpoints and critical point:

$$f(-3)=(-3-1{)}^2+3=16+3=19$$

$$f(1)=(1-1{)}^2+3=3$$

Answer:

• Absolute minimum = 3 at $x=1$

• Absolute maximum = 19 at $x=-3$

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Question 6.

Find the maximum profit that a company can make, if the profit function is given by:

$$p(x)=41-72x-18x^2$$

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Solution

Given:

$$p(x)=41-72x-18x^2$$This is a quadratic function of the form:

$$p(x)=a{x}^2+bx+c$$

where $a=-18<0$, so the parabola opens downwards → maximum exists.

To find the value of $x$ at which maximum profit occurs:

$$x=-\frac{b}{2a}$$

Here $a=-18$, $b=-72$

$$x=-\frac{-72}{2(-18)}=\frac{72}{-36}=-2$$

Now substitute $x=-2$ into profit function:

$$p(-2)=41-72(-2)-18(-2{)}^2$$

$$=41+144-18(4)$$

$$=41+144-72=113$$

Final Answer

Maximum profit = ₹ 113
Occurs when $x=-2$

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Question 7.

Find both the maximum value and the minimum value of

$$f(x)=3x^4\text{–}8x^3+12x^2\text{–}48x+25$$

on the interval $[0,3]$.

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Solution

Step 1: Differentiate

$$f(x)=3x^4-8x^3+12x^2-48x+25$$
$$f^{\mathrm{\prime }}(x)=12x^3-24x^2+24x-48$$

Factor:$$f^{\mathrm{\prime }}(x)=12(x^3-2x^2+2x-4)$$

Group:

$$x^3-2x^2+2x-4=x^2(x-2)+2(x-2)=(x^2+2)(x-2)$$

So:

$$f^{\mathrm{\prime }}(x)=12(x^2+2)(x-2)$$

Step 2: Critical points

$$f^{\mathrm{\prime }}(x)=0\Rightarrow (x^2+2)(x-2)=0$$

Since $x^2+2>0$ always, only solution is:

$$x=2$$

Step 3: Evaluate function at

• Endpoints $x=0,3$

• Critical point $x=2$

At $x=0$

$$f(0)=25$$

At $x=2$

$$f(2)=3(16)-8(8)+12(4)-48(2)+25$$

$$=48-64+48-96+25=-39$$

At $x=3$

$$f(3)=3(81)-8(27)+12(9)-48(3)+25$$

$$=243-216+108-144+25=16$$

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Question 8.

At what points in the interval $[0,2\pi ]$, does the function $\sin 2x$ attain its maximum value?

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Solution

We know that the maximum value of the sine function is 1.

So we need the values of $x$ for which:

$$\sin 2x=1$$

This happens when:

$$2x=\frac{\pi }{2}+2\pi n,\text{where }n\text{ is any integer}$$

Divide both sides by 2:

$$x=\frac{\pi }{4}+\pi n$$

Now find values in the interval $[0,2\pi ]$:

For $n=0$:

$$x=\frac{\pi }{4}$$

For $n=1$:

$$x=\frac{\pi }{4}+\pi =\frac{\pi }{4}+\frac{4\pi }{4}=\frac{5\pi }{4}$$

For $n=2$:

$$x=\frac{\pi }{4}+2\pi >2\pi \Rightarrow \text{not in interval}$$

Final Answer

$$\boxed{{\displaystyle \text{The function }\sin 2x\text{ attains its maximum value at }x=\frac{\pi }{4},\frac{5\pi }{4}}}$$

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Question 9.

What is the maximum value of the function $\sin x+\cos x$?

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Solution

We want to find the maximum value of:

$$f(x)=\sin x+\cos x$$

Use the identity:

$$\sin x+\cos x=\sqrt{2}(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)$$

Since

$$\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}=\cos \frac{\pi }{4}$$

$$\sin x+\cos x=\sqrt{2}(\sin x\sin \frac{\pi }{4}+\cos x\cos \frac{\pi }{4})$$

$$=\sqrt{2}\cos (x-\frac{\pi }{4})$$

We know:$$-1\le \cos (x-\frac{\pi }{4})\le 1$$

Multiply by $\sqrt{2}$:

$$-\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$$

Thus, the maximum value is:

$$\boxed{{\displaystyle \sqrt{2}}}$$

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Question 10.

Find the maximum value of

$$f(x)=2x^3\text{–}24x+107$$

in the interval $[1,3]$.
Find the maximum value of the same function in $[\text{–}3,\text{–}1]$.

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Solution

Given:

$$f(x)=2x^3\text{–}24x+107$$

Step 1: Differentiate

$$f^{\mathrm{\prime }}(x)=6x^2-24=6(x^2-4)=6(x-2)(x+2)$$

Step 2: Find critical points

$$f^{\mathrm{\prime }}(x)=0\Rightarrow (x-2)(x+2)=0$$

So:

$$x=2,x=-2$$

Now we will analyze each interval separately.

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Part (a): Interval $[1,3]$

Critical point inside interval = $x=2$

Evaluate at endpoints and critical point

$$f(1)=2(1{)}^3-24(1)+107=2-24+107=85$$

$$f(2)=2(8)-24(2)+107=16-48+107=75$$

$$f(3)=2(27)-24(3)+107=54-72+107=89$$

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Maximum value in $[1,3]$

$$\boxed{{\displaystyle \text{Maximum value is }89\text{ at }x=3}}$$

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Part (b): Interval $[\text{–}3,\text{–}1]$

Critical point inside interval = $x=-2$

Evaluate at endpoints and critical point

$$f(-3)=2(-27)-24(-3)+107=-54+72+107=125$$

$$f(-2)=2(-8)-24(-2)+107=-16+48+107=139$$

$$f(-1)=2(-1)-24(-1)+107=-2+24+107=129$$

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Maximum value in $[-3,-1]$

$$\boxed{{\displaystyle \text{Maximum value is }139\text{ at }x=-2}}$$