Question

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

• A. 47.66
• B. 57.66
• C. 67.66
• D. 77.66

Answer 1

Answer: (D)

77.66

We have $f\left(x\right)=3x^2+15x+5$
$\Rightarrow f^{\prime }\left(x\right)=6x+15$
Let $x=3$, and $\Delta x=.02$
Thus using approximation,
$f(x+\Delta x)\approx f\left(x\right)+f^{\prime }\left(x\right)\Delta x=(3x^2+15x+5)+(6x+15)\Delta x$
$\therefore f(3+.02)=f\left(3.02\right)\approx \left(3\right(3{)}^2+15\left(2\right)+5)+(6\left(3\right)+15\left)\right(.02)$
$=77+33\left(.02\right)=77+.66=77.66$