We are tasked with finding $k$ such that the line $y = -\frac{3}{4}x + 3$ is tangent to the graph of the function $f(x) = \frac{k}{x}$.

To do this, we need two conditions to be true at the point of tangency:

1. The y-coordinates of the function and the line must be equal at some point $x_0$ (they must intersect).
2. The slopes of the function and the line must be equal at the same point $x_0$ (they must have the same tangent slope).

Step 1: Set up the equation for intersection

At the point of tangency, the y-values of the line and the function must be equal. So, we equate the two expressions for $y$: $\frac{k}{x} = -\frac{3}{4}x + 3$ Multiply both sides of this equation by $x$ to eliminate the fraction: $k = -\frac{3}{4}x^2 + 3x$

Step 2: Set up the equation for equal slopes

The slope of the line is given as $-\frac{3}{4}$.

Next, we find the slope of the function $f(x) = \frac{k}{x}$ by calculating its derivative: $f'(x) = \frac{d}{dx} \left( \frac{k}{x} \right) = -\frac{k}{x^2}$ At the point of tangency, the derivative of the function (the slope of the tangent) must equal the slope of the line: $-\frac{k}{x_0^2} = -\frac{3}{4}$ Simplifying this: $\frac{k}{x_0^2} = \frac{3}{4}$ Multiply both sides by $x_0^2$ to solve for $k$: $k = \frac{3}{4}x_0^2$

Step 3: Substitute $k$ into the intersection equation

Now, we substitute $k = \frac{3}{4}x_0^2$ into the intersection equation from Step 1: $\frac{3}{4}x_0^2 = -\frac{3}{4}x_0^2 + 3x_0$ Add $\frac{3}{4}x_0^2$ to both sides: $2 \cdot \frac{3}{4}x_0^2 = 3x_0$ Simplify: $\frac{3}{2}x_0^2 = 3x_0$ Divide both sides by 3: $\frac{1}{2}x_0^2 = x_0$ Multiply both sides by 2: $x_0^2 = 2x_0$ Divide by $x_0$ (assuming $x_0 \neq 0$): $x_0 = 2$

Step 4: Solve for $k$

Finally, substitute $x_0 = 2$ into $k = \frac{3}{4}x_0^2$: $k = \frac{3}{4}(2^2) = \frac{3}{4} \times 4 = 3$

Thus, $k = 3$.

Conclusion:

The value of $k$ such that the line is tangent to the graph of $f(x) = \frac{k}{x}$ is $k = 3$.

Would you like more details, or do you have any questions? Here are five related problems to explore this further:

1. Find the tangent line to $f(x) = \frac{4}{x}$ at $x = 2$.
2. Determine the point where the line $y = -x + 5$ is tangent to $f(x) = \frac{2}{x}$.
3. Find the area between $f(x) = \frac{k}{x}$ and the x-axis from $x = 1$ to $x = 4$ for $k = 3$.
4. Investigate how many points of tangency exist between the graph of $f(x) = \frac{k}{x}$ and a horizontal line.
5. For what values of $k$ does the function $f(x) = \frac{k}{x}$ not have a real tangent line with slope $1$?

Tip: Always remember that at the point of tangency, both the function and its derivative must match the line’s value and slope!