Let f(x) = y = 16/x. So, f'(x) = -16/x2
Δy = f(x+Δx) - f(x) = f(3) - f(4) = 16/3 - 4 = 4/3 = 1.333
dy = f'(x)Δx = f'(x)dx = f'(4)(-1) = 16/16 = 1
Mo Y.
asked 04/16/20Compute (a) Δy and (b) dy for the given values of x and dx=Δx // y=16/x, x=4, Δx=−1 .
Δy=
dy=
Let f(x) = y = 16/x. So, f'(x) = -16/x2
Δy = f(x+Δx) - f(x) = f(3) - f(4) = 16/3 - 4 = 4/3 = 1.333
dy = f'(x)Δx = f'(x)dx = f'(4)(-1) = 16/16 = 1
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Mo Y.
Thanks a lot Mr. Mark04/17/20