Browse Questions

# If $f(x) = 3x^2 + 15x + 5$, then the approximate value of $f (3.02)$ is

$(A)\; 47.66 \quad (B)\; 57.66 \quad (C)\; 67.66 \quad (D)\; 77.66$

Toolbox:
• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
Step 1:
$x+\Delta x=3.02$
Let $x=3$
$\Delta x=0.02$
$f(x+\Delta x)=f(x)+\Delta f(x)$
$\qquad\qquad=f(x)+f'(x)\Delta x$------(1)
$f(x)=3x^2+15x+5$
Step 2:
We have $x=3$
$f(3)=3(3)^2+15(3)+5$
$\quad\;\;\;=3\times 9+45+5$
$\quad\;\;\;=27+45+5$
$\quad\;\;\;=77$
Step 3:
$f(x)=3x^2+15x+5$
$f'(x)=6x+15$
$f'(3)=6(3)+15$
$\qquad=18+15$
$\qquad=33$
Step 4:
Putting the value of $f(x)$ and $f'(x)$
$f(3.02)=77+33\times 0.02$
$\qquad\;\;\;\;=77+0.66$
$\qquad\;\;\;\;=77.66$
$f(3.02)=77.66$
Part (D) is the correct answer.