Full solution

Q. what are the coordinates of the point on the graph of

y=e^{3x}

at which the tangent line to the graph at that point also passes through the origin.

Find Derivative of $y=e^{3x}$ : First, we need to find the derivative of $y=e^{3x}$ , which will give us the slope of the tangent line at any point $x$ . $\frac{dy}{dx} = 3e^{3x}$
Calculate Tangent Line Slope: Since the tangent line passes through the origin $(0,0)$ , its slope is also the ratio of the $y$ -coordinate to the $x$ -coordinate of the point of tangency. $\newline$ So, slope = $\frac{y}{x}$
Set Derivative Equal to $y/x$ : We set the derivative equal to $\frac{y}{x}$ to find the coordinates of the point of tangency. $\newline$ $3e^{3x} = \frac{y}{x}$
Solve for y in terms of x: Now we solve for y in terms of x: $y = 3xe^{3x}$
Find Specific Point: To find the specific point, we need to solve for $x$ when the tangent line passes through the origin. This means we need to find $x$ such that the slope of the tangent line ( $3e^{3x}$ ) is equal to $y/x$ .
Find Specific Point: To find the specific point, we need to solve for $x$ when the tangent line passes through the origin. This means we need to find $x$ such that the slope of the tangent line $3e^{3x}$ is equal to $y/x$ .We can set up the equation $3e^{3x} = \frac{y}{x}$ and substitute $y$ with $3xe^{3x}$ to get: $\newline$ $3e^{3x} = \frac{3xe^{3x}}{x}$
Find Specific Point: To find the specific point, we need to solve for $x$ when the tangent line passes through the origin. This means we need to find $x$ such that the slope of the tangent line $3e^{3x}$ is equal to $y/x$ .We can set up the equation $3e^{3x} = \frac{y}{x}$ and substitute $y$ with $3xe^{3x}$ to get: $\newline$ $3e^{3x} = \frac{3xe^{3x}}{x}$ Simplify the equation: $3e^{3x} = 3e^{3x}$ $\newline$ This is true for all $x$ , so we need additional information to find the specific point. We know that the line passes through the origin, so the y-coordinate must be $x$ $0$ when $x$ is $x$ $0$ .
Find Specific Point: To find the specific point, we need to solve for $x$ when the tangent line passes through the origin. This means we need to find $x$ such that the slope of the tangent line ( $3e^{3x}$ ) is equal to $y/x$ .We can set up the equation $3e^{3x} = y/x$ and substitute $y$ with $3xe^{3x}$ to get: $\newline$ $3e^{3x} = (3xe^{3x})/x$ Simplify the equation: $3e^{3x} = 3e^{3x}$ $\newline$ This is true for all $x$ , so we need additional information to find the specific point. We know that the line passes through the origin, so the y-coordinate must be $x$ $0$ when $x$ is $x$ $0$ .However, if $x$ $3$ when $x$ $4$ , this does not satisfy the original function $x$ $5$ , since $x$ $6$ is not $x$ $0$ . We made a mistake in our assumption. We need to find a point where $x$ is not $x$ $0$ , but the slope of the tangent line is still the ratio of $y$ to $x$ .