Question

1. the functions f and g have continuous second derivatives. the table above gives values of the functions and their derivatives at selected values of x.

(a) let k(x) = f(g(x)). write an equation for the line tangent to the graph of k at x = 6.
(the table has columns x, f(x), f’(x), g(x), g’(x) with rows x=1: f(x)=-6, f’(x)=3, g(x)=2, g’(x)=8; x=2: f(x)=2, f’(x)=-2, g(x)=-3, g’(x)=0; x=3: f(x)=8, f’(x)=7, g(x)=6, g’(x)=2; x=6: f(x)=4, f’(x)=5, g(x)=3, g’(x)=-1)

Explanation:

Step1: Find \( k(6) \)

Since \( k(x) = f(g(x)) \), substitute \( x = 6 \). First, find \( g(6) \) from the table: \( g(6) = 3 \). Then \( k(6) = f(g(6)) = f(3) \). From the table, \( f(3) = 8 \). So \( k(6) = 8 \).

Step2: Find \( k'(6) \) using the Chain Rule

The Chain Rule states that \( k'(x) = f'(g(x)) \cdot g'(x) \). For \( x = 6 \), we have \( k'(6) = f'(g(6)) \cdot g'(6) \). We know \( g(6) = 3 \) and \( g'(6) = -1 \) (from the table). Then \( f'(g(6)) = f'(3) \), and from the table, \( f'(3) = 7 \). So \( k'(6) = 7 \cdot (-1) = -7 \).

Step3: Write the tangent line equation

The point - slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1) \) is a point on the line and \( m \) is the slope. For the tangent line to \( k(x) \) at \( x = 6 \), \( x_1 = 6 \), \( y_1 = k(6)=8 \), and the slope \( m = k'(6)= - 7 \). Substituting these values into the point - slope form, we get \( y - 8=-7(x - 6) \). We can also simplify this to slope - intercept form: \( y-8=-7x + 42\), so \( y=-7x + 50 \).

Answer:

The equation of the tangent line to the graph of \( k(x) \) at \( x = 6 \) is \( y - 8=-7(x - 6) \) (or \( y=-7x + 50 \))