Question

Find the approximate value of the function f(x) = `sqrt(x^2 + 3x)` at x = 1.02.

Answer 1

f(x) = `sqrt(x^2 + 3x)`

∴ f'(x) = `d/dx(sqrt(x^2 + 3x))`

= `(1)/(2sqrt(x^2 + 3x)).d/dx(x^2 + 3x)`

= `(1)/(2sqrt(x^2 + 3x)) xx (2x + 3 xx 1)`

= `(2x + 3)/(2sqrt(x^2 + 1)`
Take a = 1 and h = 0.02.

Then f(a) = f(1) = `sqrt(1^2 + 3(1)` = 2
and
f'(a) = f'(1)

= `(2(1) + 3)/(2sqrt(1^2 + 3(1)`

= `(5)/(2 xx 2)`

= `(5)/(4)`
The formula for approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ f(1.02) = f(1 + 0.02)
≑ f(1) + (0.02)f'(1)

≑ `2 + 0.02 xx(5)/(4)`

≑ `(8 + 0.1)/(4)`

= `(8.1)/(4)`
= 2.025
∴ f1.02) ≑ 2.025.