The questions are down below.

Question

Let h(x)=g(3x)

a) Find the equation of the tangent line to h(x) at x=1.

b) What is h'(0)?

c) Evaluate lim h(x)/x-2

x to 2

Answer 1

Assuming that g is differentiable at the needed points (which we'll find below), the Chain Rule gives:

h'(x) = g'(3x) * 3 = 3g'(3x)

We'll use this in all three parts.

For (a), a point on the line is given by x = 1, y = h(1) = g(3). The slope is h'(1) = 3g'(3). So assuming g is differentiable at x = 3, an equation for the tangent line (using either the linearization formula or point-slope form) is:

y = g(3) + 3g'(3)(x - 1)

(b) is less work; h'(0) = 3g'(0), assuming that g is differentiable at x = 0.

In (c), you should recognize the limit as the definition of the derivative of h(x) at x = 2. From the work above, this is h'(2) = 3g'(6), assuming g is differentiable at x = 6.