Question

Find the approximate value of f(3.02), up to 2 places of decimal, where f(x) = 3x² + 5x + 3.

Answer 1

Given: f(x) = 3x² + 5x + 3
Let x=3 and \[∆ x\] =0.02 .

Then,

\[f\left( 3 . 02 \right) = f\left( x + ∆ x \right) = 3 \left( x + ∆ x \right)^2 + 15\left( x + ∆ x \right) + 3\]

As \[∆ y = f\left( x + ∆ x \right) - f\left( x \right)\],

\[f\left( x + ∆ x \right) = f\left( x \right) + ∆ y = f\left( x \right) + f'\left( x \right) \cdot ∆ x\]

And

\[f\left( 3 . 02 \right) = 3 x^2 + 15x + 3 + \left( 6x + 15 \right) \cdot ∆ x\]

\[ = 3 \left( 3 \right)^2 + 15\left( 3 \right) + 3 + \left[ 6\left( 3 \right) + 15 \right]\left( 0 . 02 \right)\]

\[ = 27 + 45 + 3 + \left( 33 \right)\left( 0 . 02 \right) \]

\[ = 75 + 0 . 66\]

\[ = 75 . 66\]