Find the average value of the function over the given interval and all values z

Question

Find the average value of the function over the given interval and all values z in the interval for which the function equals its average value. f(z) =x^2+2/z^2, 1 le z le 7 Use a graphing utility to verify your results. The average is 9/14 and the point at which the function is equal to its mean value is The average is 9/7 and the point at which the function is equal to its mean value is The average is 9/14 and the point at which the function is equal to its mean value is squareroot7-squareroot7 the average is 9/7 and the point at which the function is equal to its mean value is squareroot7 -squareroot7 The average is 9/14 and the point at which the function is equal to its mean value is -squareroot7

Explanation / Answer

f(z) = (z2 +2)/z2 , 1 <= z <= 7

Average value = 1/(7 -1) [1 to 7] (z2 +2)/z2 dz

= (1/6) [1 to 7] 1 + 2z-2 dz

= (1/6) [1 to 7] z - 2/z                     since xn dx = xn+1/(n +1)

= (1/6) [7 - 2/7 - ( 1 - 2/1) ] = (1/6) [54/7] = 9/7

Hence average value = 9/7

9/7 = (z2 +2)/z2

==> 9z2 = 7(z2 +2)

==> 9z2 - 7z2 = 14

==> z2 = 7 ==> z = +7 , -7

but z is defined in the interval [1 ,7]

Hence z = 7

Therefore average value = 9/7 and point at which function equals mean value =7 (Option B)

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