Questions: Differentiate the function. Then find an equation of the tangent line at the indicated point on the graph of the function. w=g(z)=3+sqrt(16-z), (z, w)=(15,4) The derivative of the function w=g(z)=3+sqrt(16-z) is -(1/(2 sqrt(16-z))). (Type an exact answer, using radicals as needed.) An equation of the tangent line is (Type an equation.)

Solution Steps

To solve this problem, we need to first find the derivative of the function \( w = g(z) = 3 + \sqrt{16 - z} \). This will give us the slope of the tangent line at any point \( z \). Then, we evaluate this derivative at the given point \( z = 15 \) to find the slope of the tangent line at that specific point. Finally, we use the point-slope form of a line to find the equation of the tangent line at the point \( (15, 4) \).

To find the derivative of the function \( w = g(z) = 3 + \sqrt{16 - z} \), we apply the chain rule. The derivative is:

\[ g'(z) = -\frac{1}{2\sqrt{16 - z}} \]

Substitute \( z = 15 \) into the derivative to find the slope of the tangent line at this point:

\[ g'(15) = -\frac{1}{2\sqrt{16 - 15}} = -\frac{1}{2} \]

The point-slope form of a line is given by:

\[ w - w_1 = m(z - z_1) \]

where \( m \) is the slope, and \((z_1, w_1)\) is the point on the line. Here, \( m = -\frac{1}{2} \), \( z_1 = 15 \), and \( w_1 = 4 \). Substituting these values, we get:

\[ w - 4 = -\frac{1}{2}(z - 15) \]

Simplifying, the equation of the tangent line is:

\[ w = -\frac{1}{2}z + \frac{15}{2} + 4 \]

\[ w = -\frac{1}{2}z + \frac{23}{2} \]

Final Answer