Originally Posted by
Soroban Hello, math34!
Evidently, we have the function: .$\displaystyle f(x) \:=\:x^3$
Then: .$\displaystyle df \:=\:3x^2dx$
We know that: .$\displaystyle f(8) \,= \,8^3 \,= \,512$
Let $\displaystyle x = 8,\;dx = -0.2$
Then: .$\displaystyle dy \:=\:3(8^2)(-0.2)\:=\:-38.4$
Hence: .$\displaystyle 7.8^3\:\approx\:512 - 38.4 \:=\:\boxed{473.6}$
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How good is this approximation?
The actual value is: .$\displaystyle 7.8^3\:=\:474.552$
Our answer is short by only: .$\displaystyle 474.552 - 473.6 \:=\:0.952$
As a percent: .$\displaystyle \frac{0.952}{474.552}\:=\:0.002066103 \:\approx\:0.2\%$
We were off by only two-tenths of a percent (2 parts in 1000) . . . not bad!