Find the area of the region bounded by the curve y = f(x) = (x^3)-4x+1 and the tangent line to the curve y = f(x) at (-1,4).
A. 9/4
B. 8.25
C. 27/4
D. 31/4
E. 6.25
To find the area of the region bounded by the curve and the tangent line, we need to determine the x-values where the curve and the tangent line intersect.
First, let's find the equation of the tangent line to the curve y = (x^3)-4x+1 at the point (-1,4). The slope of the tangent line is given by the derivative of the function f(x) = (x^3)-4x+1 evaluated at x = -1.
f'(x) = 3x^2 - 4
f'(-1) = 3(-1)^2 - 4 = -1 - 4 = -5
The equation of the tangent line is therefore y = -5x + c, where c is the y-intercept. We can find c by substituting the coordinates of the point (-1,4) into the equation.
4 = -5(-1) + c
4 = 5 + c
c = -1
So the equation of the tangent line is y = -5x - 1.
Next, we need to find the x-values where the curve y = (x^3)-4x+1 intersects with the tangent line y = -5x - 1.
Setting the y-values equal to each other, we have:
(x^3)-4x+1 = -5x - 1
Rearranging and simplifying the equation:
x^3 + x - 2x = -2 - 1
x^3 - x = -3
We can see that x = -1 is a solution to this equation. By polynomial division, we can find the other solutions. Dividing x^3 - x by x - (-1) using synthetic division:
-1 | 1 0 -1
| -1 1
| 1 -1 0
Therefore, x^3 - x = (x + 1)(x^2 - x) = 0
The other two solutions are x = 0 and x = 1.
So, the curve y = (x^3)-4x+1 intersects the tangent line y = -5x - 1 at three points: (-1, 4), (0, 1), and (1, -6).
To find the area of the region bounded by the curve and the tangent line, we integrate the difference of the two functions between the x-values of 0 and 1.
Area = ∫[0,1] [(x^3)-4x+1 - (-5x - 1)] dx
Simplifying the integrand:
Area = ∫[0,1] (x^3 + x + 2x + 2) dx
= ∫[0,1] (x^3 - 9x + 2) dx
Integrating term by term:
Area = [1/4 * x^4] - [9/2 * x^2] + [2 * x] |_[0,1]
= [1/4 * (1)^4] - [9/2 * (1)^2] + [2 * (1)] - [1/4 * (0)^4] - [9/2 * (0)^2] + [2 * (0)]
Simplifying:
Area = 1/4 - 9/2 + 2 - 0 - 0 + 0
= (1/4) - (9/2) + 2
= 1/4 - 18/4 + 8/4
= -17/4 + 8/4
= -9/4
Since area cannot be negative, we take the absolute value:
Area = 9/4
Therefore, the area of the region bounded by the curve and the tangent line is 9/4.
The correct answer is A. 9/4.
To find the area of the region bounded by the curve and its tangent line, we need to find the points of intersection between the curve and the tangent line.
First, let's find the equation of the tangent line to the curve y = f(x) at (-1,4).
To find the slope of the tangent line, we can take the derivative of the function f(x).
f'(x) = 3x^2 - 4
Now, let's find the slope of the tangent line at x = -1.
f'(-1) = 3(-1)^2 - 4 = 3 - 4 = -1
Therefore, the slope of the tangent line at (-1,4) is -1.
Now, let's find the equation of the tangent line using the point-slope form.
y - y1 = m(x - x1)
y - 4 = -1(x - (-1))
y - 4 = -x - 1
y = -x + 3
Next, we need to find the points of intersection between the curve y = f(x) and the tangent line y = -x + 3.
Setting the two equations equal to each other, we get:
(x^3) - 4x + 1 = -x + 3
Rearranging the equation, we have:
x^3 - 4x + x - 3 + 1 = 0
x^3 - 3x - 2 = 0
By synthetic division, we can find that x = 1 is a root of the equation.
(x - 1)(x^2 + x + 2) = 0
Setting (x^2 + x + 2) = 0, we can solve for x using the quadratic formula.
x = (-1 ± √(-7))/(2)
Since we are looking for real solutions, we can eliminate the negative square root value.
x = (-1 + √(-7))/(2)
Since the square root is negative, there are no real solutions for x, meaning the curve and the tangent line do not intersect.
This means that the area of the region bounded by the curve and the tangent line is 0.
Therefore, the answer is not provided in the given options.