Let f(x) = x^3 - x^2 + x - 1. Find an equation for tangent line to f at x = 2.
x=2 could be rewritten as (2,0)
Look at example 1:
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f(x) = x^3 - x^2 + x - 1
f'(x) = 3x^2 - 2x + 1
when x=2 , f'(1) = 12 - 4 + 1 = 9
also when x=2, f(2) = 8 - 4 + 2 -1 = 5
so the tangent has a slope of 9 and passes through (2,5)
equation of tangent is y = 9x + b
at (2.5)
5 = 9(2) + b
b = -13
equation of tangent: y = 9x - 13
To find the equation of the tangent line to the function f(x) at a given point, we need to find the slope of the tangent line and the coordinates of the point.
First, we find the slope of the tangent line by taking the derivative of the function f(x). The derivative of f(x) = x^3 - x^2 + x - 1 can be found by differentiating each term separately.
f'(x) = 3x^2 - 2x + 1
Next, we evaluate the derivative at the given point x = 2 to find the slope of the tangent line.
f'(2) = 3(2)^2 - 2(2) + 1
= 12 - 4 + 1
= 9
Now, we have the slope of the tangent line, which is 9.
To find the coordinates of the point on the graph, we substitute the given x-value into the function f(x).
f(2) = (2)^3 - (2)^2 + 2 - 1
= 8 - 4 + 2 - 1
= 5
Therefore, the coordinates of the point on the graph are (2, 5).
Using the point-slope form of a linear equation, which is y - y1 = m(x - x1), we can find the equation of the tangent line.
y - 5 = 9(x - 2)
Simplifying this equation, we get:
y - 5 = 9x - 18
Finally, we can rewrite this equation in slope-intercept form, y = mx + b, where m represents the slope and b is the y-intercept:
y = 9x - 13
Therefore, the equation for the tangent line to f(x) at x = 2 is y = 9x - 13.