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What is the equation of the tangent line of #f(x)=(x-1)^3 # at #x=2#?

Answer 1

Charles Anctil

y = 3x - 5

To find the equation in the form y = mx + c , where m represents the gradient and c , the y-intercept.

Given an x-coordinate, f(2) is the value of the y-coordinate and f'(2) is the value of the tangent gradient.

differentiate using the #color(blue)" chain rule " #

# d/dx [ f(g(x)) ] = f'(g(x)) . g'(x) #

# f'(x) = 3(x-1)^2 .d/dx(x-1) = 3(x-1)^2 #

and f'(2) =#3(2-1)^2 = 3 = " m gradient of tangent " #

The "partial" equation is therefore y = 3x + c.

now # f(2) = (3-2)^2 = 1 → (2,1)" is point on tangent " #

y = 3x + c using (2,1): 6 + c = 1 → c = -5

and the tangent equation is thus: y = 3x - 5.

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Answer 2

Scarlett Allison

The equation of the tangent line of f(x)=(x-1)^3 at x=2 is y = 9x - 17.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.