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Use the first derivative test to find and classify all local extrema in the interval
(0,4) for the function

f(x)=-3x^(4)ln(3x)

$15$ . Use the first derivative test to find and classify all local extrema in the interval $(0,4)$ for the function $\newline$ $f(x)=-3 x^{4} \ln (3 x)$

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Find derivatives of exponential functions

Full solution

Q.

15

. Use the first derivative test to find and classify all local extrema in the interval

(0,4)

for the function

\newline

f(x)=-3 x^{4} \ln (3 x)

Find First Derivative: To find the local extrema, we first need to find the first derivative of the function $f(x) = -3x^{4}\ln(3x)$ . Using the product rule for differentiation, where if $f(x) = u(x)v(x)$ , then $f'(x) = u'(x)v(x) + u(x)v'(x)$ , we can find the derivative of $f(x)$ . Let $u(x) = -3x^{4}$ and $v(x) = \ln(3x)$ . Then we need to find $u'(x)$ and $v'(x)$ .
Apply Product Rule: First, we find $u'(x)$ . Since $u(x) = -3x^{4}$ , the derivative $u'(x)$ is found by bringing down the exponent and reducing the exponent by one. $\newline$ $u'(x) = -3 \times 4x^{4-1} = -12x^{3}.$
Find $u'(x)$ and $v'(x)$ : Next, we find $v'(x)$ . Since $v(x) = \ln(3x)$ , we use the chain rule for differentiation, which states that if $v(x) = \ln(g(x))$ , then $v'(x) = \frac{g'(x)}{g(x)}$ . Here, $g(x) = 3x$ , so $g'(x) = 3$ . Therefore, $v'(x) = \frac{3}{3x} = \frac{1}{x}$ .
Calculate $f'(x)$ : Now we can find $f'(x)$ using the product rule. $\newline$ $f'(x) = u'(x)v(x) + u(x)v'(x)$ $\newline$ $= (-12x^{3}) \cdot \ln(3x) + (-3x^{4}) \cdot (\frac{1}{x})$ $\newline$ $= -12x^{3}\ln(3x) - 3x^{3}$ .
Find Critical Points: To find the critical points, we set $f'(x)$ to zero and solve for $x$ . $\newline$ $0 = -12x^{3}\ln(3x) - 3x^{3}$ $\newline$ $0 = x^{3}(-12\ln(3x) - 3).$
Solve for x: Since $x^{3}$ is never zero in the interval $(0,4)$ , we can divide both sides by $x^{3}$ to find the critical points. $\newline$ $0 = -12\ln(3x) - 3$ $\newline$ $12\ln(3x) = -3$ $\newline$ $\ln(3x) = -\frac{1}{4}.$
Check Interval: To solve for $x$ , we exponentiate both sides of the equation. $e^{\ln(3x)} = e^{(-1/4)}$ $3x = e^{(-1/4)}.$
Classify Critical Point: Now we solve for $x$ . $x = \frac{e^{(-\frac{1}{4})}}{3}$ .
Use First Derivative Test: We need to check if this critical point is in the interval $(0,4)$ . Since $e^{(-1/4)}$ is a positive number less than $1$ , and we are dividing by $3$ , the critical point $x = e^{(-1/4)} / 3$ is indeed in the interval $(0,4)$ .
Choose Test Points: To classify the critical point, we use the first derivative test. We need to check the sign of $f'(x)$ to the left and right of the critical point $x = e^{-1/4} / 3$ .
Evaluate $f'(x_1)$ : Choose test points $x_1$ and $x_2$ such that $0 < x_1 < e^{-1/4} / 3 < x_2 < 4$ and evaluate $f'(x_1)$ and $f'(x_2)$ . Let's choose $x_1 = e^{-1/4} / 6$ and $x_2 = e^{-1/4} / 2$ .
Evaluate $f'(x_2)$ : Evaluate $f'(x_1) = -12x_1^{3}\ln(3x_1) - 3x_1^{3}$ . $\newline$ Since $\ln(3x_1)$ is negative (because $3x_1 < 1$ ) and $x_1^{3}$ is positive, $f'(x_1)$ will be positive.
Evaluate $f'(x_2)$ : Evaluate $f'(x_1) = -12x_1^{3}\ln(3x_1) - 3x_1^{3}$ . Since $\ln(3x_1)$ is negative (because $3x_1 < 1$ ) and $x_1^{3}$ is positive, $f'(x_1)$ will be positive.Evaluate $f'(x_2) = -12x_2^{3}\ln(3x_2) - 3x_2^{3}$ . Since $\ln(3x_2)$ is negative (because $3x_2 < e$ ) and $x_2^{3}$ is positive, $f'(x_2)$ will be positive.

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