if the

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 $x=3, \Delta x=0.02]$

$=(27+45+5)+(18+15)(0.02)$

$=77+(33)(0.02)$

$=77+0.66$

$=77.66$

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

The correct answer is D.