Let 

​f(x)equals=2424StartRoot x EndRootxminus−4 x4x.

**a.** Find all points on the graph of f at which the tangent line is horizontal.

**b.** Find all points on the graph of f at which the tangent line has slope 

negative five halves−52.

Question content area bottom

Part 1

The tangent line is horizontal at the​ point(s) 

left parenthesis 9 comma 36 right parenthesis(9,36).

​(Simplify your answer. Type an ordered pair. Use a comma to separate answers as​ needed.)

Part 2

The tangent line has slope of 

negative five halves−52 

at the​ point(s) 

enter your response here.

​(Simplify your answer. Type an ordered pair. Use a comma to separate answers as​ needed.)

Let's start by analyzing the function \( f(x) = 24 \sqrt{x} - 4x \).

### Part 1: Finding Points with a Horizontal Tangent Line

**Step 1:** Calculate the first derivative of \( f(x) \):
\[
f(x) = 24 \sqrt{x} - 4x = 24x^{1/2} - 4x
\]
Using the power rule:
\[
f'(x) = 24 \cdot \frac{1}{2} x^{-1/2} - 4 = 12 x^{-1/2} - 4 = \frac{12}{\sqrt{x}} - 4
\]

**Step 2:** Set the first derivative equal to zero to find where the tangent line is horizontal:
\[
\frac{12}{\sqrt{x}} - 4 = 0
\]

**Step 3:** Solve for \( x \):
\[
\frac{12}{\sqrt{x}} = 4
\]
\[
12 = 4 \sqrt{x}
\]
\[
\sqrt{x} = 3
\]
\[
x = 9
\]

**Step 4:** Find the corresponding \( y \)-value by substituting \( x \) back into the original function \( f(x) \):
\[
f(9) = 24 \sqrt{9} - 4(9) = 24 \cdot 3 - 36 = 72 - 36 = 36
\]

So, the point on the graph where the tangent line is horizontal is:
\[
(9, 36)
\]

### Part 2: Finding Points with a Tangent Line Slope of \(-\frac{5}{2}\)

**Step 1:** Set the first derivative equal to \(-\frac{5}{2}\):
\[
\frac{12}{\sqrt{x}} - 4 = -\frac{5}{2}
\]

**Step 2:** Solve for \( x \):
\[
\frac{12}{\sqrt{x}} = -\frac{5}{2} + 4
\]
Convert 4 to a fraction:
\[
\frac{12}{\sqrt{x}} = -\frac{5}{2} + \frac{8}{2} = \frac{3}{2}
\]

**Step 3:** Solve for \( x \):
\[
\frac{12}{\sqrt{x}} = \frac{3}{2}
\]
\[
24 = 3 \sqrt{x}
\]
\[
\sqrt{x} = 8
\]
\[
x = 64
\]

**Step 4:** Find the corresponding \( y \)-value by substituting \( x \) back into the original function \( f(x) \):
\[
f(64) = 24 \sqrt{64} - 4(64) = 24 \cdot 8 - 256 = 192 - 256 = -64
\]

So, the point on the graph where the tangent line has a slope of \(-\frac{5}{2}\) is:
\[
(64, -64)
\]

### Summary:
- The tangent line is horizontal at the point \((9, 36)\).
- The tangent line has a slope of \(-\frac{5}{2}\) at the point \((64, -64)\).

Would you like more details or have any questions? Here are some related questions for further exploration:

1. How do you find the second derivative of \( f(x) \) and what does it tell you about the concavity of the function?
2. What are the critical points of the function \( f(x) \)?
3. How do you determine if the critical points are local minima, maxima, or saddle points?
4. What is the inflection point of \( f(x) \)?
5. How can you graph the function \( f(x) \) to visualize these points?
6. What is the significance of the discriminant in solving equations?
7. How do changes in the coefficients of \( f(x) \) affect the critical points and points of horizontal tangency?
8. How would you solve the same problem using a graphing calculator or computer software?

**Tip:** When solving for points where the tangent line has a specific slope, always ensure to substitute the critical \( x \)-values back into the original function to find the corresponding \( y \)-values accurately.

Explore how to find points on the graph of f(x) where the tangent line is horizontal or has a specific slope using differentiation techniques. Detailed step-by-step solutions provided.

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