Question

Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ x=2 \\ x+∆x=2.01 \\ \Rightarrow ∆x=0.01 \\ f\left(x\right)=4x^2+5x+2 \\ \Rightarrow f\left(x=2\right)=16+10+2=28 \\ \mathrm{Now},y=f\left(x\right) \\ \Rightarrow \frac{dy}{dx}=8x+5 \\ \therefore dy=∆y=\frac{dy}{dx}dx=\left(8x+5\right)\times 0.01=\left(16+5\right)\times 0.01=0.21 \\ \therefore f\left(2.01\right)=y+∆y=28.21 \end{gathered}$$