The equation of the tangent line to f(x) at the point (2,6) is y=6x-6.?

Question

the equation of the tangent line to g(x) at the point (2,3) is y=y=2x-1. The function h(x) is f(x)/g(x). what is the equation of the tangent line to h(x) at the point (2,3)

Answer 1

The tangent line to the function f(x) is y = 6x - 6 and the point (2, 6).

The tangent line to the function g(x) is y = 2x - 1 and the point (2, 3).

The function h(x) = f(x)/g(x) and the point is (2, 3).

Differentiate with respect to x.

h'(x) = [g(x) * f'(x) - f(x) * g'(x)]/[g(x)]²

Substitute x = 2 in the above equation

h'(2) = [g(2) * f'(2) - f(2) * g'(2)]/[g(2)]²

f'(x) = 6x - 6 ⇒ f'(2) = 6(2) - 6 = 6.

g'(x) = 2x - 1 ⇒ g'(2) = 2(2) - 1 = 3.

Find f(x):

ʃ f'(x) dx = ʃ (6x - 6) dx

f(x) = 3x² - 6x + C

Find the value of C, substitute x = 2, and f(x) = 3 in the above function.

3 = 3*2² - 6*2 + C ⇒ C = 3

f(x) = 3x² - 6x + 3 ⇒ f(2) = 3*2² - 6*2 + 3 = 3.

Find g(x):

ʃ g'(x) dx = ʃ (2x - 1) dx

g(x) = x² - x + C

Find the value of C, substitute x = 2, and g(x) = 3 in the above function.

3 = 2² - 2 + C ⇒ C = 1

g(x) = x² - x + 1 ⇒ g(2) = 2² - 2 + 1 = 3.

h'(2) = [g(2) * f'(2) - f(2) * g'(2)]/[g(2)]²

h'(2) = [3 * 6 - 3 * 3]/3² = 1.

Substitute m = 1 and (x₁, y₁) = (2,3) in the point-slope form of line equation: y - y₁ = m(x - x₁).

y - 3  = 1(x - 2)

y  = x + 1

The tangent line to function h(x) at (2, 3) is y  = x + 1.