Tangent to f at x=2 is y =4x+1 and tangent to g at x=2 is y=3x-2.

Question

Find the line tangent to the following curves at x=2.?

a. y=f(x)g(x)
b. y=f(x)/g(x)

 

Answer 1

(b).

The tangent line to the function f(x) at x = 2 is y = 4x + 1.

The y-value of the point is y = 4(2)+1 = 9.

The point is (2, 9).

The tangent line to the function g(x) at x = 2 is y = 3x - 2.

The y-value of the point is y = 3(2)-2 = 4.

The point is (2, 4).

Find the tangent line to the function y = f(x)/g(x) at x = 2.

Differentiate with respect to x.

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

Substitute x = 2 in the above equation

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

f'(x) = 4x+1 ⇒ f'(2) = 4(2)+1 = 9.

g'(x) = 3x - 2 ⇒ g'(2) = 3(2) - 2 = 4.

Find f(x):

ʃ f'(x) dx = ʃ (4x+1) dx

f(x) = 2x² + x + C

Find the value of C, substitute x = 2, and f(x) = 9 in the above function.

9 = 2*2² + 2 + C ⇒ C = -1

f(x) = 2x² + x + 1 ⇒ f(2) = 2*2² + 2 + 1 = 11.

Find g(x):

ʃ g'(x) dx = ʃ (3x - 2) dx

g(x) = (3/2)x² - 2x + C

Find the value of C, substitute x = 2, and g(x) = 4 in the above function.

4 = (3/2)*2² - 2*2 + C ⇒ C = 2

g(x) = (3/2)x² - 2x + 2 ⇒ g(2) = (3/2)*2² - 2*2 + 2 = 4.

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

y'(2) = [4 * 9 - 11 * 4]/4² = -1/2.

To find y-value, substitute x = 2 in the function y = f(x)/g(x).

y = f(2)/g(2) = 11/4.

Substitute m = - 1/2 and (x₁, y₁) = (2, 11/4) in the point-slope form of line equation: y - y₁ = m(x - x₁).

y - 11/4  = -1/2(x - 2)

2(y - 11/4)  = -1(x - 2)

2y - 11/2  = -x + 2

2y = -x + 2 + 11/2

2y = -x + 15/2

y = -x/2 + 15/4

The tangent line to function y = f(x)/g(x) at (2, 11/4) is y = -x/2 + 15/4.

 

Answer 2

(a).

The tangent line to the function f(x) at x = 2 is y = 4x + 1.

The y-value of the point is y = 4(2)+1 = 9.

The point is (2, 9).

The tangent line to the function g(x) at x = 2 is y = 3x - 2.

The y-value of the point is y = 3(2)-2 = 4.

The point is (2, 4).

Find the tangent line to the function y = f(x)*g(x) at x = 2.

Differentiate with respect to x.

y'(x) = f(x) * g'(x) + g(x) * f'(x)

Substitute x = 2 in the above equation

y'(2) = f(2) * g'(2) + g(2) * f'(2)

f'(x) = 4x+1 ⇒ f'(2) = 4(2)+1 = 9.

g'(x) = 3x - 2 ⇒ g'(2) = 3(2) - 2 = 4.

Find f(x):

ʃ f'(x) dx = ʃ (4x+1) dx

f(x) = 2x² + x + C

Find the value of C, substitute x = 2, and f(x) = 9 in the above function.

9 = 2*2² + 2 + C ⇒ C = -1

f(x) = 2x² + x + 1 ⇒ f(2) = 2*2² + 2 + 1 = 11.

Find g(x):

ʃ g'(x) dx = ʃ (3x - 2) dx

g(x) = (3/2)x² - 2x + C

Find the value of C, substitute x = 2, and g(x) = 4 in the above function.

4 = (3/2)*2² - 2*2 + C ⇒ C = 2

g(x) = (3/2)x² - 2x + 2 ⇒ g(2) = (3/2)*2² - 2*2 + 2 = 4.

Find the slope of tangent line to function y = f(x)*g(x) at x = 2.

y'(2) = f(2) * g'(2) + g(2) * f'(2)

y'(2) = 11 * 4 + 4 * 9 = 80.

To find y-value, substitute x = 2 in the function y = f(x)*g(x).

y = f(2)*g(2) = 11*4 = 44.

Substitute m = 80 and (x₁, y₁) = (2, 44) in the point-slope form of line equation: y - y₁ = m(x - x₁).

y - 44 = 80(x - 2)

y - 44 = 80x - 160

y = 80x - 116

The tangent line to function y = f(x)*g(x) at (2, 44) is y = 80x - 116.