Question

If $f\left(x\right)=3x^2+15x+5,$ then the approximate value of f(3.02) is ?
(a) 47.66 (b) 57.66 (c) 67.66 (d) 77.66

Answer 1

Consider $f\left(x\right)=3x^2+15x+5\Rightarrow f^{\prime }\left(x\right)=6x+15$
Let x = 3 and $\Delta x=0.02$ Also, $f(x+\Delta x)\simeq f\left(x\right)+\Delta x{f}^{\prime }\left(x\right)$.
$\Rightarrow f(x+\Delta x)\simeq (3x^2+15x+5)+(6x+15)\Delta x$
$\Rightarrow f\left(3.02\right)\simeq 3\times 3^2+15\times 3+5+(6\times 3+15)\left(0.02\right)$ (as x = 3, $\Delta x=0.02$)
= 27 + 45 + 5 + (18 + 15)(0.02) = 77 + 33(0.02) = 77 + 0.66
$\Rightarrow f\left(3.02\right)\simeq 77.66$
Hence, the approximate value of f(3.02) is 77.66. $\therefore$ The correct option is (d).