|Let f(0) = 3 and f'(0) = 2. Then the equation of the tangent line to the graph of y = f(x) at x = 0 is y =

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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### Problem Statement

**Given:**
- \( f(0) = 3 \)
- \( f'(0) = 2 \)

### Question:
Determine the equation of the tangent line to the graph of \( y = f(x) \) at \( x = 0 \).

---

### Solution:
To find the equation of the tangent line, we use the point-slope form of a line:

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

Where:
- \( (x_1, y_1) \) is a point on the line. Here, \( x_1 = 0 \) and \( y_1 = f(0) = 3 \).
- \( m \) is the slope of the line, which is \( f'(0) = 2 \).

Plugging in these values, we get:

\[ y - 3 = 2(x - 0) \]

Simplifying this, the equation of the tangent line is:

\[ y = 2x + 3 \]

### Answer:
\[ y = 2x + 3 \]
Transcribed Image Text:### Problem Statement **Given:** - \( f(0) = 3 \) - \( f'(0) = 2 \) ### Question: Determine the equation of the tangent line to the graph of \( y = f(x) \) at \( x = 0 \). --- ### Solution: To find the equation of the tangent line, we use the point-slope form of a line: \[ y - y_1 = m(x - x_1) \] Where: - \( (x_1, y_1) \) is a point on the line. Here, \( x_1 = 0 \) and \( y_1 = f(0) = 3 \). - \( m \) is the slope of the line, which is \( f'(0) = 2 \). Plugging in these values, we get: \[ y - 3 = 2(x - 0) \] Simplifying this, the equation of the tangent line is: \[ y = 2x + 3 \] ### Answer: \[ y = 2x + 3 \]
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