Find the approximate value of (3.98)^3. - Vidyadip

Question

Find the approximate value of $(3.98)^3$.

✓

Answer

Let $f(x)=x^3$
Differentiate w.r. t. $x$.
$f^{\prime}(x)=3 x^2$
Let $a=4, h=-0.02$
For $x=a=4$, from (I) we get
$f(a)=f(4)=(4)^3=64$
For $x=a=4$, from (II) we get
$f^{\prime}(a)=f^{\prime}(4)=3(4)^2=48$
We have, $f(a+h) \doteqdot f(a)+h f^{\prime}(a)$
$
\begin{array}{r}
f[4+(-0.02)] \doteqdot f(4)+(-0.02) \cdot f^{\prime}(4) \\
f(3.98) \doteqdot 64+(-0.02) .(48) \quad \ldots
\end{array}
$
[From (III) and (IV)]
$
\begin{aligned}
& f(3.98) \doteqdot 64-0.96 \\
& \therefore f(3.98)=(3.98)^3 \doteqdot 63.04 \\
&
\end{aligned}
$

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