Question

Find the approximate value of $f\left(5.001\right)$, where $f\left(x\right)=x^3-7x^2+15$.

Answer 1

Let $x=5$ and $\Delta x=0.001$
Now, $\Delta y=f(x+\Delta x)-f\left(x\right)$
$\therefore f(x+\Delta x)=f\left(x\right)+\Delta y$
$\approx f\left(x\right)+f^{\prime }\left(x\right)\cdot \Delta x$ (as $dx=\Delta x$)
$\therefore f\left(5.001\right)\approx (x^3-7x^2+15)+(3x^2-14x)\Delta x$
$=\left[\right(5{)}^3-7(5{)}^2+15]+\left[3\right(5{)}^2-14\left(5\right)\left]\right(0.001)$
$=(125+175+15)+(75-70)\left(0.001\right)$
$=-35+\left(5\right)\left(0.001\right)=-35+0.005=-34.995$
Hence, the approximate value of $f\left(5.001\right)$ is $34.995$