If the normal to the curve y = f (x) at the point (3, 4) makes an angle `3pi//4` with the positive x-axis, then f' (3) =

Options

A. `-1`
B. `-3/4`
C. `4/3`
D. 1

Answer

To solve the problem, we need to find the derivative \( f'(3) \) given that the normal to the curve \( y = f(x) \) at the point \( (3, 4) \) makes an angle of \( \frac{3\pi}{4} \) with the positive x-axis.

### Step 1: Understanding the slope of the normal
The angle \( \theta = \frac{3\pi}{4} \) corresponds to the slope of the normal line. The slope \( m \) of the normal can be calculated using the tangent of the angle:
\[ m = \tan\left(\frac{3\pi}{4}\right) \]
The tangent of \( \frac{3\pi}{4} \) is:
\[ \tan\left(\frac{3\pi}{4}\right) = -1 \]
Thus, the slope of the normal line is:
\[ m = -1 \]

### Step 2: Relating the slope of the normal to the derivative
The slope of the normal line is related to the derivative of the function at the point of tangency. If \( f'(3) \) is the slope of the tangent line at \( x = 3 \), then the relationship between the slopes is given by:
\[ m_{\text{normal}} = -\frac{1}{f'(3)} \]
Since we found that \( m_{\text{normal}} = -1 \), we can set up the equation:
\[ -1 = -\frac{1}{f'(3)} \]

### Step 3: Solving for \( f'(3) \)
Now, we can solve for \( f'(3) \):
\[ -1 = -\frac{1}{f'(3)} \implies 1 = \frac{1}{f'(3)} \implies f'(3) = 1 \]

### Final Answer
Thus, the value of \( f'(3) \) is:
\[ \boxed{1} \]