Question

question 21 · 1 point if $q(x)=0.6x^{3}3^{x}$ and $a = 3$, what is $q(a)$ to the nearest tenth? (do not include \$q(3)=\ in your answer.) provide your answer below:

Explanation:

Step1: Apply product - rule for differentiation

If $q(x)=0.6x\cdot3^{x}$, then $q^{\prime}(x)=0.6(3^{x}+x\cdot3^{x}\ln(3))$

Step2: Substitute $x = a = 3$

$q^{\prime}(3)=0.6(3^{3}+3\cdot3^{3}\ln(3))$
$q^{\prime}(3)=0.6(27 + 81\ln(3))\approx0.6(27+81\times1.0986)$
$q^{\prime}(3)=0.6(27 + 88.9866)=0.6\times115.9866 = 69.59196\approx69.6$

Answer:

$69.6$