Let g(x)=(ln(x))/(x). What is the absolute maximum value of g ? Choose 1 answer: (A) -e (B) (1)/(e) (C) -(1)/(e^(2)) (D) g has no maximum value

Find Critical Points: To find the absolute maximum value of the function g(x) = (ln(x))/x, we need to find the critical points of the function. This involves taking the derivative of g(x) with respect to x and setting it equal to zero.Calculate Derivative: The derivative of g(x) with respect to x is g'(x) = (1/x * x - ln(x) * 1)/(x^2) by using the quotient rule, which is (d(u/v)/dx = (v * du/dx - u * dv/dx)/v^2 where u = ln(x) and v = x).Solve for Critical Points: Simplifying the derivative, we get g'(x) = (1 - ln(x))/(x^2).Determine Maximum: To find the critical points, we set g'(x) equal to zero and solve for x: (1 - ln(x))/(x^2) = 0.Evaluate at Critical Point: This gives us the equation 1 - ln(x) = 0, which simplifies to ln(x) = 1.Taking the exponential of both sides to solve for x, we get e^(ln(x)) = e^1, which simplifies to x = e.Now we need to determine if this critical point corresponds to a maximum, minimum, or neither. We can do this by using the second derivative test or by evaluating the function at the critical point and comparing to the behavior at the endpoints.Since g(x) is only defined for x > 0, and as x approaches infinity, g(x) approaches 0, we can infer that the critical point at x = e is likely to be a maximum. To confirm, we can evaluate g(x) at x = e.Substituting x = e into g(x), we get g(e) = (ln(e))/e = 1/e.Since g(x) approaches 0 as x approaches infinity and is undefined for x ≤ 0, the value of g(e) = 1/e is the absolute maximum value of g(x) on its domain (0, ∞).

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Let
g(x)=(ln(x))/(x).
What is the absolute maximum value of
g ?
Choose 1 answer:
(A)
-e
(B)
(1)/(e)
(C)
-(1)/(e^(2))
(D)
g has no maximum value

Let $g(x)=\frac{\ln (x)}{x}$ . $\newline$ What is the absolute maximum value of $g$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-e$ $\newline$ (B) $\frac{1}{e}$ $\newline$ (C) $-\frac{1}{e^{2}}$ $\newline$ (D) $g$ has no maximum value

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Let

g(x)=\frac{\ln (x)}{x}

\newline

What is the absolute maximum value of

g

\newline

Choose

1

answer:

\newline

(A)

-e

\newline

(B)

\frac{1}{e}

\newline

(C)

-\frac{1}{e^{2}}

\newline

(D)

g

has no maximum value

Find Critical Points: To find the absolute maximum value of the function $g(x) = \frac{\ln(x)}{x}$ , we need to find the critical points of the function. This involves taking the derivative of $g(x)$ with respect to $x$ and setting it equal to zero.
Calculate Derivative: The derivative of $g(x)$ with respect to $x$ is $g'(x) = \frac{1}{x} \cdot x - \ln(x) \cdot 1)/(x^2)$ by using the quotient rule, which is $\frac{d(\frac{u}{v})}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$ where $u = \ln(x)$ and $v = x)$ .
Solve for Critical Points: Simplifying the derivative, we get $g'(x) = \frac{1 - \ln(x)}{x^2}$ .
Determine Maximum: To find the critical points, we set $g'(x)$ equal to zero and solve for $x$ : $\frac{1 - \ln(x)}{x^2} = 0$ .
Evaluate at Critical Point: This gives us the equation $1 - \ln(x) = 0$ , which simplifies to $\ln(x) = 1$ .
Evaluate at Critical Point: This gives us the equation $1 - \ln(x) = 0$ , which simplifies to $\ln(x) = 1$ . Taking the exponential of both sides to solve for $x$ , we get $e^{\ln(x)} = e^1$ , which simplifies to $x = e$ .
Evaluate at Critical Point: This gives us the equation $1 - \ln(x) = 0$ , which simplifies to $\ln(x) = 1$ . Taking the exponential of both sides to solve for $x$ , we get $e^{\ln(x)} = e^1$ , which simplifies to $x = e$ . Now we need to determine if this critical point corresponds to a maximum, minimum, or neither. We can do this by using the second derivative test or by evaluating the function at the critical point and comparing to the behavior at the endpoints.
Evaluate at Critical Point: This gives us the equation $1 - \ln(x) = 0$ , which simplifies to $\ln(x) = 1$ . Taking the exponential of both sides to solve for $x$ , we get $e^{\ln(x)} = e^1$ , which simplifies to $x = e$ . Now we need to determine if this critical point corresponds to a maximum, minimum, or neither. We can do this by using the second derivative test or by evaluating the function at the critical point and comparing to the behavior at the endpoints. Since $g(x)$ is only defined for x > 0, and as $x$ approaches infinity, $g(x)$ approaches $0$ , we can infer that the critical point at $x = e$ is likely to be a maximum. To confirm, we can evaluate $g(x)$ at $x = e$ .
Evaluate at Critical Point: This gives us the equation $1 - \ln(x) = 0$ , which simplifies to $\ln(x) = 1$ . Taking the exponential of both sides to solve for $x$ , we get $e^{\ln(x)} = e^1$ , which simplifies to $x = e$ . Now we need to determine if this critical point corresponds to a maximum, minimum, or neither. We can do this by using the second derivative test or by evaluating the function at the critical point and comparing to the behavior at the endpoints. Since $g(x)$ is only defined for x > 0, and as $x$ approaches infinity, $g(x)$ approaches $0$ , we can infer that the critical point at $x = e$ is likely to be a maximum. To confirm, we can evaluate $g(x)$ at $x = e$ . Substituting $x = e$ into $g(x)$ , we get $\ln(x) = 1$ $5$ .
Evaluate at Critical Point: This gives us the equation $1 - \ln(x) = 0$ , which simplifies to $\ln(x) = 1$ . Taking the exponential of both sides to solve for $x$ , we get $e^{\ln(x)} = e^1$ , which simplifies to $x = e$ . Now we need to determine if this critical point corresponds to a maximum, minimum, or neither. We can do this by using the second derivative test or by evaluating the function at the critical point and comparing to the behavior at the endpoints. Since $g(x)$ is only defined for x > 0, and as $x$ approaches infinity, $g(x)$ approaches $0$ , we can infer that the critical point at $x = e$ is likely to be a maximum. To confirm, we can evaluate $g(x)$ at $x = e$ . Substituting $x = e$ into $g(x)$ , we get $\ln(x) = 1$ $5$ . Since $g(x)$ approaches $0$ as $x$ approaches infinity and is undefined for $\ln(x) = 1$ $9$ , the value of $x$ $0$ is the absolute maximum value of $g(x)$ on its domain $x$ $2$ .