Question

Suppose there exists a function, f(x), such that f(1) = 5 and f'(1) = 4. Let h(x) = f(x)/x. Find the equation of the tangent line to h(x) at x = 1.

Answer 1

To find the equation of the tangent line to h(x) at x = 1, we need to find the derivative of h(x) and evaluate it at x = 1. The slope of the tangent line is -1, and the equation of the tangent line is y = -x + 6.

Step-by-step explanation:

To find the equation of the tangent line to h(x) at x = 1, we need to find the derivative of h(x) and evaluate it at x = 1. First, let's find the derivative of h(x):

h(x) = f(x)/x

Using the quotient rule, the derivative of h(x) is:

h'(x) = (f'(x) * x - f(x)) / x^2

Substituting f(1) = 5, f'(1) = 4, and x = 1 into h'(x), we get:

h'(1) = (4*1 - 5) / 1^2 = -1

Therefore, the slope of the tangent line to h(x) at x = 1 is -1. The equation of a line with slope -1 passing through the point (1, h(1)) can be found using the point-slope form:

y - h(1) = -1(x - 1)

Since h(1) = f(1)/1 = 5, the equation of the tangent line to h(x) at x = 1 is:

y - 5 = -1(x - 1)

y - 5 = -x + 1

y = -x + 6