write the equation of line tangent to f(x)=(1/3x^3)-x^2+x-1 when x=1
I meant to write the equation like this: f(x)=(1/3)x^3-x^2+x-1 when x=1
(1/3x^3)-x^2+x-1 when x=1
F'(x) = x^2 -2x + 1
F'(1) = 1-2+ 1 = 0
f(x) = 1/3-1+ 1 -1
f(x) =( 1-3+3-3)/3
f(x) = -2/3
y - (-2/3 ) = 0(x -1)
y + 2/3 = 0
y = -2/3
To find the equation of the line tangent to the function f(x) = (1/3)x^3 - x^2 + x - 1 at the point where x = 1, we need to follow these steps:
Step 1: Find the derivative of the function f(x) using differentiation.
The derivative of f(x) can be found by applying the power rule of differentiation for each term of the function. For this function, we have:
f'(x) = d/dx[(1/3)x^3] - d/dx[x^2] + d/dx[x] - d/dx[1]
= (1/3)(3x^2) - 2x + 1 - 0
= x^2 - 2x + 1
Step 2: Substitute x = 1 into f'(x) to find the slope of the tangent line.
f'(1) = (1)^2 - 2(1) + 1
= 1 - 2 + 1
= 0
The slope of the tangent line is 0.
Step 3: Find the y-coordinate of the point where x = 1 on the function f(x).
To find the y-coordinate, plug x = 1 into the original function f(x):
f(1) = (1/3)(1)^3 - (1)^2 + (1) - 1
= 1/3 - 1 + 1 - 1
= 1/3 - 1/3
= 0
The point where x = 1 on the function f(x) is (1, 0).
Step 4: Write the equation of the tangent line using the slope-intercept form.
Since the slope of the tangent line is 0 and it passes through the point (1, 0), the equation of the line can be written as:
y = 0
In this case, the tangent line will be a horizontal line passing through the point (1, 0).