if the


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

A. 47.66 B. 57.66 C. 67.66 D. 77.66


Let $x=3$ and $\Delta x=0.02$. Then, we have:

$f(3.02)=f(x+\Delta x)=3(x+\Delta x)^{2}+15(x+\Delta x)+5$

Now, $\Delta y=f(x+\Delta x)-f(x)$

$\begin{aligned} \Rightarrow f(x+\Delta x) &=f(x)+\Delta y \\ & \approx f(x)+f^{\prime}(x) \Delta x \end{aligned}$       $($ As $d x=\Delta x)$

$\Rightarrow f(3.02) \approx\left(3 x^{2}+15 x+5\right)+(6 x+15) \Delta x$

$=\left[3(3)^{2}+15(3)+5\right]+[6(3)+15](0.02) \quad[$ As $x=3, \Delta x=0.02]$





Hence, the approximate value of f(3.02) is 77.66.

The correct answer is D.

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