Given that $y=\frac{e^{2 x}}{x}$ for $x>0$ , find the range of values of $x$ for which $y$ is decreasing.

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Q. Given that

y=\frac{e^{2 x}}{x}

for

x>0

, find the range of values of

x

for which

y

is decreasing.

Find Decreasing Points: To find where $y$ is decreasing, we need to find where its derivative $y'$ is negative.
Apply Quotient Rule: Differentiate $y$ using the quotient rule: $y' = \frac{v(u') - u(v')}{v^2}$ , where $u = e^{2x}$ and $v = x$ .
Calculate Derivatives: Calculate $u'$ (derivative of $e^{2x}$ ): $u' = 2e^{2x}$ .
Simplify Result: Calculate $v'$ (derivative of $x$ ): $v' = 1$ .
Set Condition for Negativity: Apply the quotient rule: $y' = \frac{x(2e^{2x}) - e^{2x}(1)}{x^2}$ .
Identify Sign Dependence: Simplify $y'$ : $y' = \frac{2xe^{2x} - e^{2x}}{x^2}$ .
Solve for Negative Range: Factor out $e^{2x}$ : $y' = \frac{e^{2x}(2x - 1)}{x^2}$ .
Correct Math Error: Set $y'$ less than $0$ to find where $y$ is decreasing: $\frac{e^{2x}(2x - 1)}{x^2} < 0$ .
Correct Math Error: Set $y'$ less than $0$ to find where $y$ is decreasing: $\frac{e^{2x}(2x - 1)}{x^2} < 0$ . Since $e^{2x}$ is always positive and $x^2$ is always positive for $x > 0$ , the sign of $y'$ depends on $(2x - 1)$ .
Correct Math Error: Set $y'$ less than $0$ to find where $y$ is decreasing: $\frac{e^{2x}(2x - 1)}{x^2} < 0$ . Since $e^{2x}$ is always positive and $x^2$ is always positive for $x > 0$ , the sign of $y'$ depends on $(2x - 1)$ . Set $(2x - 1) < 0$ to find when $y'$ is negative: $0$ $1$ .
Correct Math Error: Set $y'$ less than $0$ to find where $y$ is decreasing: $\frac{e^{2x}(2x - 1)}{x^2} < 0$ . Since $e^{2x}$ is always positive and $x^2$ is always positive for $x > 0$ , the sign of $y'$ depends on $(2x - 1)$ . Set $(2x - 1) < 0$ to find when $y'$ is negative: $0$ $1$ . Solve for $0$ $2$ : $0$ $3$ .
Correct Math Error: Set $y'$ less than $0$ to find where $y$ is decreasing: $\frac{e^{2x}(2x - 1)}{x^2} < 0$ . Since $e^{2x}$ is always positive and $x^2$ is always positive for $x > 0$ , the sign of $y'$ depends on $(2x - 1)$ . Set $(2x - 1) < 0$ to find when $y'$ is negative: $0$ $1$ . Solve for $0$ $2$ : $0$ $3$ . Divide by $0$ $4$ : $0$ $5$ .
Correct Math Error: Set $y'$ less than $0$ to find where $y$ is decreasing: $\frac{e^{2x}(2x - 1)}{x^2} < 0$ . Since $e^{2x}$ is always positive and $x^2$ is always positive for $x > 0$ , the sign of $y'$ depends on $(2x - 1)$ . Set $(2x - 1) < 0$ to find when $y'$ is negative: $0$ $1$ . Solve for $0$ $2$ : $0$ $3$ . Divide by $0$ $4$ : $0$ $5$ . However, we have a math error because we forgot that $0$ $2$ must be greater than $0$ , so the range of $0$ $2$ for which $y$ is decreasing is $y$ $0$ .