Question

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

Answer 1

\[\text { Let }: \]

\[ x = 2\]

\[x + ∆ x = 2 . 01\]

\[ \Rightarrow ∆ x = 0 . 01\]

\[f\left( x \right) = 4 x^2 + 5x + 2\]

\[ \Rightarrow f\left( x = 2 \right) = 16 + 10 + 2 = 28\]

\[\text { Now,} y = f\left( x \right)\]

\[ \Rightarrow \frac{dy}{dx} = 8x + 5\]

\[ \therefore dy = ∆ y = \frac{dy}{dx}dx = \left( 8x + 5 \right) \times 0 . 01 = \left( 16 + 5 \right) \times 0 . 01 = 0 . 21\]

\[ \therefore f\left( 2 . 01 \right) = y + ∆ y = 28 . 21\]