https://file.aiquickdraw.com/mathbot/file/17270349566532r39fjyi.jpg

Let's break down and solve this problem step-by-step.

We are given \( f(x) = 4x - x^2 \), and we need to find the slope of the tangent line at the point \( (1, 3) \) using two different definitions of the derivative.

### Part (a) - Using the Definition of the Derivative:
We are given two forms of the derivative definition:

1. Using the limit definition:
\[
f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}
\]
For this, \( a = 1 \) and \( f(x) = 4x - x^2 \).

Let's compute:

- \( f(1) = 4(1) - (1)^2 = 4 - 1 = 3 \)
- \( f(1 + h) = 4(1 + h) - (1 + h)^2 = 4 + 4h - (1 + 2h + h^2) = 4 + 4h - 1 - 2h - h^2 = 3 + 2h - h^2 \)

Now substitute into the limit:
\[
f'(1) = \lim_{h \to 0} \frac{(3 + 2h - h^2) - 3}{h} = \lim_{h \to 0} \frac{2h - h^2}{h} = \lim_{h \to 0} (2 - h) = 2
\]

So, the slope of the tangent line is \( f'(1) = 2 \).

2. Using the alternate limit form:
\[
f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
\]
Here, we use \( a = 1 \) again.

- \( f(1) = 3 \)
- \( f(x) = 4x - x^2 \)

Now apply the limit:
\[
f'(1) = \lim_{x \to 1} \frac{(4x - x^2) - 3}{x - 1} = \lim_{x \to 1} \frac{4x - x^2 - 3}{x - 1}
\]
Simplifying the numerator:
\[
4x - x^2 - 3 = -(x - 1)(x - 3)
\]
So:
\[
f'(1) = \lim_{x \to 1} \frac{-(x - 1)(x - 3)}{x - 1} = \lim_{x \to 1} -(x - 3) = -2
\]

It seems there is a discrepancy here. Let me verify both expressions.Both forms of the derivative yield the same result:

- Using the limit definition: \( f'(1) = 2 \)
- Using the alternate limit form: \( f'(1) = 2 \)

### Part (b) - Equation of the Tangent Line:
Now that we know the slope of the tangent line is \( 2 \) and the point of tangency is \( (1, 3) \), we can use the point-slope form of a line:

\[
y - y_1 = m(x - x_1)
\]

Substitute \( m = 2 \), \( x_1 = 1 \), and \( y_1 = 3 \):

\[
y - 3 = 2(x - 1)
\]

Simplifying:

\[
y = 2x + 1
\]

So, the equation of the tangent line is \( y = 2x + 1 \).

### Summary:
- The slope of the tangent line at \( (1, 3) \) is \( 2 \).
- The equation of the tangent line is \( y = 2x + 1 \).

Would you like any more details on the steps? Here are five related questions you could explore next:

1. How does the definition of a derivative relate to the concept of instantaneous rate of change?
2. Can you find the tangent line equation for other points on the curve \( f(x) = 4x - x^2 \)?
3. What is the geometric interpretation of the tangent line at a specific point on a curve?
4. How do higher-order derivatives relate to the curvature of the graph?
5. What happens to the tangent line as \( x \) approaches infinity for this function?

**Tip:** Always double-check your limits when computing derivatives to avoid simple mistakes in factorization or simplification.

Let f(x) = 4x - x^2. Find the slope of the tangent line to the graph of f at the point (1, 3) using the definition of the derivative. Use the results to write an equation for the tangent line.

This problem involves finding the slope of the tangent line to the graph of f(x) = 4x - x^2 at the point (1, 3) using two forms of the derivative. Detailed step-by-step explanations and key concepts such as limits, derivatives, and the point-slope form of a line are included. Suitable for students in Grades 11-12 studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation