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Find the absolute maximum and the absolute minimum of f(x) = 8ze^2t for x in the integral [-3, 3] Find the average value of the function f(x) = 300/(4x + 1)^2 on the interval [0, 2].

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5 (1 Ratings )

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**Question**

Find the absolute maximum and the absolute minimum of f(x) = 8ze^2t for x in the integral [-3, 3] Find the average value of the function f(x) = 300/(4x + 1)^2 on the interval [0, 2].

** **

**Solution**

9) we have given f(x)=8xe2x for x in [-3,3]

f'(x)=8[2xe2x+e2x]

f'(x)=8[2xe2x+e2x] =0 implies e2x(2x+1)=0 implies x=-1/2 and e2x =0 implies 2x=e0=1 implies x=1/2

x=-1/2 and x=1/2

At x=-1/2

f(-1/2) =8*(-1/2)*e2*(-1/2)=(-4)*e-1=-1.471

At x=1/2

f(1/2) =8*(1/2)*e2*(1/2)=4*e =10.873

At x=-3

f(-3) =8*(-3)*e2*(-3)=(-24)e-6=-0.059

At x=3

f(3) =8*(3)*e2*(3)=(24)e6= 9682.291

Absolute maximum is 9682.291 at x=3 for given function

Absolute minimum is -1.471 at x=-1/2 for given function

10) we have given f(x)=300/(4x+1)2 on [0,2]

we know the average value of function is =1/(b-a)*integration of (x=a to b)f(x)dx

we have given a=0 and b=2

average value of f(x) =1/(2-0)*integration of (x=0 to 2)(300/(4x+1)2)dx

=(1/2)*integration of (x=0 to 2)(300/(4x+1)2)dx

substitute u=4x+1 implies du=4dx implies dx=du/4

=(1/2)*integration of (x=0 to 2)(300/(u)2)(du/4)

=(75/2)*integration of (x=0 to 2)(u-2)du

=(75/2)*[u-1/(-1)] from x=0 to 2

=(75/2)*[-(1/u)] from x=0 to 2

=(75/2)*[-(1/(4x+1))] from x=0 to 2 since u=4x+1

=(75/2)*{[-(1/(4*2+1))]-[-(1/(4*0+1))]}

=(75/2)*[(-1/9)+1]=(75/2)*(8/9) =(75*4)/9 =100/3

Average value of the given function is 100/3 = 33.33

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