calculus question

Question

Assume f and g are differentiable functions with ​h(x)=​f(g(x)). Suppose the equation of the line tangent to the graph of g at the point ​(1​, 7​) is y=3x+4 and the equation of the line tangent to the graph of f at ​(7​,​8) is y=-4x+36

a. Calculate ​h(1​) and h'​(​1).

b. Determine an equation of the line tangent to the graph of h at the point on the graph where x=1

Answer 1

The formula for slopes of tangent lines to g(x) is given as y = 3x + 4. This means g'(x) = 3x+4. The general antiderivative is determined as g(x) = (3/2)x² + 4x + C. Since the graph passes through (1,7) the value of C is found as:

7 = (3/2)(1)² + 4(1) + C

14 = 3 + 8 + 2C

C = 3/2

g(x) = (3/2)x² + 4x + (3/2)

Following the same pattern for f(x):

f'(x) = -4x + 36

f(x) = -2x² + 36x + K

8 = -2(7)² + 36(7) + K

K = 8 + 98 - 252

K = -146

f(x) = -2x² + 36x - 146

h(x) = f[g(x)]

h(1) = f[g(1)] = f(7) = -98 + 252 - 146 = 8

The point (1,8) is on the graph of h(x).

To find h'(1).

h(x) = f[g(x)]

h'(x) = f'[g(x)][g'(x)] (chain rule)

h'(1) = f'(7)g'(1) = (8)(7) = 56 [g(1) = 7]

That means if you graph h(x) the slope of the tangent line at (1,8) will be 56.

Equation of the tangent line to h(x) at (1,8): y = 56(x-1) + 8

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