Question

Find the approximate value of f(3.01) when f(x)=3x22+5x+4

Answer 1

The approximate value of f(3.01) is "46.23".

Step-by-step explanation:

We have,

$f(x)=3x^{2} +5x+4$

Let x = 3 and Δx =0.01

∴ $f^{1} (x) = 6x+5$

$\delta y = f^{1} (x) \deltax =(6x+5)\times  0.01$

Also,

$Δy =f(x+\delta x)-f(x)$

$f(x+\delta x)=f(x)+\delta y$$f(3.01)=(3x^{2} +5x+4)+(6x+5)(0.01)$

Putting$x,\delta xand\delta y$,

$f(3.01)=(3(3)^{2} +5(3)+4)+(6(3)+5)(0.01)$

= $(27 +15+4)+23(0.01)$

=$46+0.23$

= 46.23

Hence,  the approximate value of f(3.01) is "46.23".