If $f(x) =3x^2 +10x-5,$ then by using the concept of differentials, the approximate value of $f(2.03)$ is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 27.66
$2.03=2+0.03$
$x=2$
$Δx=0.03$
so $f(x.Δx)=f(x)+\left.\frac{df}{dx}\right]_x(Δx)$
so $f(2.03)=f(2)+(6x+10)_{x=2}(0.03)$
$=27+22×0.03$
$=27.66$