find k that such that the line is tangent to the graph of the function. f(x)=k/x, y=-(3/4)x+3


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!

find k such that the line is tangent to the graph of the function f(x)=k/x, y=-(3/4)x+3

This problem focuses on finding the value of k such that the line y = -(3/4)x + 3 is tangent to the function f(x) = k/x. It covers key calculus concepts, including tangency conditions, derivatives, and solving quadratic equations. Suitable for high school students in Grades 11-12.

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