Find the tangent line

Question

Can someone help me understand tangent lines? Here is a question I'm struggling with. Thanks.

Derivative of the function g(x) 2x^3-4x is G(x)=6x^2-4

Find the equation of the tangent line to the fuction g(x) at x=1

Answer 1

Given g(x) = 2x^3-4x

g'(x) = 6x^2-4

For x = 1, g(x) = y

y = 2x^3-4x

g(1) = 2-4 = -2

For x = 1, g'(x) = m

6x^2-4 = 6-4 = 2

So the point is (1,-2) and m = 2

line equationis y-y1 = m(x-x1)

y-(-2) = 2(x-1)

y+2 = 2x-2

y = 2x

 

 

Comments

Tangent line equation is y = 2x - 4 .

Answer 2

The function g(x) = 2x³- 4x

y = 2x³- 4x

Tangent line to the function g(x) at x = 1.

• We know the x value of the point, we need to find y value of the point

Since the tangent line touches curve g(x) at x = 1 also point on a curve.

 y = 2x³- 4x

 y = 2(1)³- 4(1)

y = 2 - 4 = - 2

y = - 2

• The derivative of the curve is equal to the slope of the tangent line.

dy/dx = 6x² - 4

m = 6x² - 4

at x = 1

m = 6(1)² - 4

m = 6 - 4

Slope of the line (m) = 2

• To find the tangent line equation, substitute the values of m = 2 and (x₁, y₁) = ( 1, - 2)  in point slope form of a line y - y₁ = m(x - x₁)

y  + 2 = 2(x - 1)

y + 2 = 2x - 2

y = 2x - 2 - 2

y = 2x - 4

The tangent line equation is y = 2x - 4.