Question

question: if $g(x)=\frac{0.3(9^{x})}{x^{3}}$, find $g(5)$ to the nearest tenth. (do not include \$g(5)=$\ in your answer.) provide your answer below:

Explanation:

Step1: Apply quotient - rule

$g(x)=\frac{0.3(9^{x})}{x^{3}}$, quotient rule: $(\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}$, where $u = 0.3(9^{x})$, $u'=0.3(9^{x})\ln(9)$, $v = x^{3}$, $v' = 3x^{2}$.

Step2: Calculate $g'(x)$

$g'(x)=\frac{0.3(9^{x})\ln(9)\cdot x^{3}-0.3(9^{x})\cdot3x^{2}}{x^{6}}$.

Step3: Simplify $g'(x)$

$g'(x)=0.3(9^{x})\frac{x^{3}\ln(9)- 3x^{2}}{x^{6}}=0.3(9^{x})\frac{x\ln(9)-3}{x^{4}}$.

Step4: Substitute $x = 5$

$g'(5)=0.3(9^{5})\frac{5\ln(9)-3}{5^{4}}\approx1314.9$.

Answer:

$1314.9$