How do you find the equation of the line tangent to #f(x) = (x-1)^3# at the point where x=2?

Answer 1

First, you have to differentiate #f(x)# using the chain rule.

Let #y = u^3# and #u = x - 1#.
#y' = 3u^2# and #u' = 1#
#dy/dx = 3u^2 xx 1#
#dy/dx = 3(x - 1)^2#

Next, we have to find the point through which our tangent, and our function pass.

#f(2) = (2 - 1)^3#
#f(2) = 1#
So, the function and the tangent pass through the point #(2, 1)#.
Now, we need to find the slope of the tangent by plugging in and evaluating #x = a# inside the derivative.
#m_"tangent" = 3(2 - 1)^2#
#m_"tangent" = 3#

We must next determine the equation of the tangent using point-slope form.

#y - y_1 = m(x - x_1)#
#y - 1 = 3(x - 2)#
#y - 1 = 3x - 6#
#y = 3x - 5#
Hence, the equation of the tangent line to #f(x) = (x - 1)^3# at the point #x= 2# is #y = 3x - 5#.

Hopefully this helps!

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Answer 2

To find the equation of the line tangent to the function f(x) = (x-1)^3 at the point where x=2, we need to find the slope of the tangent line and the coordinates of the point of tangency.

First, we find the derivative of the function f(x) using the power rule. The derivative of f(x) = (x-1)^3 is f'(x) = 3(x-1)^2.

Next, we substitute x=2 into the derivative to find the slope of the tangent line at x=2. f'(2) = 3(2-1)^2 = 3.

Now, we have the slope of the tangent line, which is 3. To find the coordinates of the point of tangency, we substitute x=2 into the original function f(x). f(2) = (2-1)^3 = 1.

Therefore, the point of tangency is (2, 1).

Using the slope-intercept form of a linear equation, y = mx + b, where m is the slope and b is the y-intercept, we can substitute the values we found to get the equation of the tangent line.

Plugging in the slope (m=3) and the coordinates of the point of tangency (x=2, y=1), we have y = 3x - 5.

Thus, the equation of the line tangent to f(x) = (x-1)^3 at the point where x=2 is y = 3x - 5.

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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.

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.

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.

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.

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