Asked by Alaina
calculate delta y for f(x)= x^(3/2) with x= 4 and delta x= dx=0.1
delta y= f(x+delta x)- f(x)
= f(4.1) - f(4)
delta y= f(4.1)-8
f(x + delta x)= 8 + 3/2*x^(1/2)*0.1
= 8 + 0.15x^(1/2)
= 8 + 0.30
= 8.30
delta y= 8.30 - 8
delta y= 0.30
but this is wrong. why is it wrong?
delta y= f(x+delta x)- f(x)
= f(4.1) - f(4)
delta y= f(4.1)-8
f(x + delta x)= 8 + 3/2*x^(1/2)*0.1
= 8 + 0.15x^(1/2)
= 8 + 0.30
= 8.30
delta y= 8.30 - 8
delta y= 0.30
but this is wrong. why is it wrong?
Answers
Answered by
Steve
Hmm. Looks good to me. But you've done a lot of extra work. Using differentials, we approximate delta y = dy
dy = f'(x) dx
= 3/2 x^(1/2) dx
= 3/2 (2) (.1)
= 0.3
dy = f'(x) dx
= 3/2 x^(1/2) dx
= 3/2 (2) (.1)
= 0.3
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