IF the lines y=-4x+b are tangents to the curve `y=1/x`, then b=

Options

A. `pm4`
B. `pm2`
C. `pm1`
D. `pm8`

Answer

To find the value of \( b \) such that the line \( y = -4x + b \) is tangent to the curve \( y = \frac{1}{x} \), we will follow these steps:

### Step 1: Find the derivative of the curve
The curve given is \( y = \frac{1}{x} \). To find the slope of the tangent line at any point on this curve, we need to compute the derivative:

\[ \frac{dy}{dx} = -\frac{1}{x^2} \]

### Step 2: Set the derivative equal to the slope of the tangent line
The slope of the line \( y = -4x + b \) is \( -4 \). For the line to be tangent to the curve, the slope of the curve at the point of tangency must equal the slope of the line:

\[ -\frac{1}{x^2} = -4 \]

### Step 3: Solve for \( x \)
To solve for \( x \), we can eliminate the negative signs and rearrange the equation:

\[ \frac{1}{x^2} = 4 \]

Taking the reciprocal of both sides gives:

\[ x^2 = \frac{1}{4} \]

Taking the square root of both sides, we find:

\[ x = \pm \frac{1}{2} \]

### Step 4: Find the corresponding \( y \) values
Now we need to find the \( y \) values corresponding to \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \):

1. For \( x = \frac{1}{2} \):

\[ y = \frac{1}{\frac{1}{2}} = 2 \]

2. For \( x = -\frac{1}{2} \):

\[ y = \frac{1}{-\frac{1}{2}} = -2 \]

### Step 5: Use the points to find \( b \)
Now we have two points of tangency: \( \left(\frac{1}{2}, 2\right) \) and \( \left(-\frac{1}{2}, -2\right) \).

We will substitute these points into the line equation \( y = -4x + b \) to find \( b \).

1. For the point \( \left(\frac{1}{2}, 2\right) \):

\[ 2 = -4\left(\frac{1}{2}\right) + b \]
\[ 2 = -2 + b \]
\[ b = 4 \]

2. For the point \( \left(-\frac{1}{2}, -2\right) \):

\[ -2 = -4\left(-\frac{1}{2}\right) + b \]
\[ -2 = 2 + b \]
\[ b = -4 \]

### Conclusion
The values of \( b \) for which the line \( y = -4x + b \) is tangent to the curve \( y = \frac{1}{x} \) are:

\[ b = 4 \quad \text{and} \quad b = -4 \]