find the equation for the derivitive, tangent, and normal line of x=-7 for g(x)=sqrt(2-x)


Let's break down the problem step by step.

### 1. Derivative of \( g(x) = \sqrt{2 - x} \)
First, we'll find the derivative of the function \( g(x) \).

#### Step 1: Rewrite \( g(x) \)
\( g(x) = \sqrt{2 - x} \) can be rewritten as:
\[
g(x) = (2 - x)^{1/2}
\]

#### Step 2: Differentiate \( g(x) \)
Now, apply the chain rule:
\[
g'(x) = \frac{1}{2}(2 - x)^{-1/2} \cdot (-1)
\]
Simplify:
\[
g'(x) = -\frac{1}{2\sqrt{2 - x}}
\]
So the derivative of \( g(x) \) is:
\[
g'(x) = -\frac{1}{2\sqrt{2 - x}}
\]

### 2. Slope of the Tangent Line at \( x = -7 \)
Now, substitute \( x = -7 \) into \( g'(x) \) to find the slope of the tangent line:
\[
g'(-7) = -\frac{1}{2\sqrt{2 - (-7)}} = -\frac{1}{2\sqrt{9}} = -\frac{1}{6}
\]
Thus, the slope of the tangent line at \( x = -7 \) is \( -\frac{1}{6} \).

### 3. Point on the Curve at \( x = -7 \)
Next, substitute \( x = -7 \) into the original function \( g(x) \) to find the corresponding point on the curve:
\[
g(-7) = \sqrt{2 - (-7)} = \sqrt{9} = 3
\]
So the point is \( (-7, 3) \).

### 4. Equation of the Tangent Line
The equation of a tangent line is given by:
\[
y - y_1 = m(x - x_1)
\]
Where:
- \( m = -\frac{1}{6} \) (slope of the tangent),
- \( (x_1, y_1) = (-7, 3) \) (point on the curve).

Substitute these values:
\[
y - 3 = -\frac{1}{6}(x - (-7)) = -\frac{1}{6}(x + 7)
\]
Simplify:
\[
y - 3 = -\frac{1}{6}x - \frac{7}{6}
\]
\[
y = -\frac{1}{6}x + \frac{18}{6} - \frac{7}{6}
\]
\[
y = -\frac{1}{6}x + \frac{11}{6}
\]
Thus, the equation of the tangent line is:
\[
y = -\frac{1}{6}x + \frac{11}{6}
\]

### 5. Equation of the Normal Line
The slope of the normal line is the negative reciprocal of the tangent line's slope. So, the slope of the normal line is:
\[
m_{\text{normal}} = 6
\]
Using the point \( (-7, 3) \), the equation of the normal line is:
\[
y - 3 = 6(x - (-7)) = 6(x + 7)
\]
Simplify:
\[
y - 3 = 6x + 42
\]
\[
y = 6x + 45
\]
Thus, the equation of the normal line is:
\[
y = 6x + 45
\]

### Summary:
- The derivative is \( g'(x) = -\frac{1}{2\sqrt{2 - x}} \).
- The equation of the tangent line at \( x = -7 \) is \( y = -\frac{1}{6}x + \frac{11}{6} \).
- The equation of the normal line at \( x = -7 \) is \( y = 6x + 45 \).

Would you like further details or explanations?

#### 5 Related Questions:
1. How is the chain rule applied to find derivatives of composite functions?
2. Why is the slope of the normal line the negative reciprocal of the tangent line's slope?
3. What is the significance of the derivative in finding the slope of the tangent line?
4. How do we generalize the process of finding tangent and normal lines for other functions?
5. How can we determine if a function is differentiable at a given point?

#### Tip:
The normal line is always perpendicular to the tangent line, meaning their slopes multiply to \(-1\).

find the equation for the derivative, tangent, and normal line of x=-7 for g(x)=sqrt(2-x)

Learn how to find the derivative, tangent line, and normal line of the function g(x) = sqrt(2 - x) at x = -7. This detailed solution includes applying the chain rule for differentiation, finding the slope of the tangent and normal lines, and forming their respective equations. Suitable for Grades 11-12 calculus students.

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