The equation of a curve is y = x^3 + x^2 - x -1
(a) Determine:
(i) the stationary points of the curve;
(ii) the nature of the stationary points in (a)(i) above.
(b) Determine:
(i) the equation of the tangent to the curve at x = 1;
(ii) the equation of the normal to the curve at x = 1
y' = 3x^2+2x-1 = (3x-1)(x+1)
y" = 6x+2 = 3(2x+1)
(a)
(i) y'=0
(ii) max if y" < 0, min if y" > 0
(b)
(i) y(1) = 0, y'(1) = 4; use point-slope form
(ii) -1/y' = -1/4
(a) (i) To find the stationary points of the curve, we need to find the points where the derivative of the function is equal to zero.
The derivative of the given equation is:
dy/dx = 3x^2 + 2x - 1
To find the stationary points, we set dy/dx equal to zero and solve for x:
3x^2 + 2x - 1 = 0
This is a quadratic equation, and we can solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 2, and c = -1. Plugging these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4(3)(-1))) / (2(3))
Simplifying this expression:
x = (-2 ± √(4 + 12)) / 6
x = (-2 ± √16) / 6
x = (-2 ± 4) / 6
This gives us two possible values for x:
x1 = (-2 + 4) / 6 = 2 / 6 = 1/3
x2 = (-2 - 4) / 6 = -6 / 6 = -1
So, the x-values of the stationary points are 1/3 and -1.
(ii) To determine the nature of the stationary points, we examine the second derivative of the function. Note that the second derivative represents the concavity of the curve.
The second derivative of the given equation is:
d^2y/dx^2 = 6x + 2
Substituting the x-values of the stationary points into the second derivative:
d^2y/dx^2 at x = 1/3:
6(1/3) + 2 = 2 + 2 = 4
d^2y/dx^2 at x = -1:
6(-1) + 2 = -6 + 2 = -4
Since the second derivative is positive at x = 1/3 and negative at x = -1, we can conclude the following about the nature of the stationary points:
- At x = 1/3, the stationary point is a local minimum (concave up).
- At x = -1, the stationary point is a local maximum (concave down).
(b) (i) To find the equation of the tangent to the curve at x = 1, we need to find the slope of the tangent line at that point.
The slope of the tangent line is equal to the derivative of the function at x = 1:
dy/dx at x = 1:
3(1)^2 + 2(1) - 1 = 3 + 2 - 1 = 4
So, the slope of the tangent line at x = 1 is 4.
Next, we use the point-slope formula to find the equation of the tangent line. We can choose any point on the curve at x = 1, for example, the y-coordinate at x = 1.
Using the original equation of the curve, substitute x = 1 to find the corresponding y-coordinate:
y at x = 1:
(1)^3 + (1)^2 - (1) - 1 = 1 + 1 - 1 - 1 = 0
So, the point (1, 0) lies on the tangent line.
Using the point-slope formula, the equation of the tangent line is:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is a point on the line. Plugging in the values:
y - 0 = 4(x - 1)
Simplifying:
y = 4x - 4
Therefore, the equation of the tangent to the curve at x = 1 is y = 4x - 4.
(ii) To find the equation of the normal to the curve at x = 1, we need to find the slope of the normal line at that point.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is -1/4.
Using the point-slope formula, the equation of the normal line is:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is a point on the line. Plugging in the values:
y - 0 = (-1/4)(x - 1)
Simplifying:
y = -1/4x + 1/4
Therefore, the equation of the normal to the curve at x = 1 is y = -1/4x + 1/4.
To solve this problem, we'll need to find the derivatives of the given curve, determine the stationary points, and then analyze the nature of those points. We'll also need to find the equations of the tangent and normal lines at a specific point.
(a) (i) To find the stationary points, we need to find the values of x where the derivative of the curve is equal to zero. The derivative of y = x^3 + x^2 - x - 1 can be found by differentiating each term separately:
dy/dx = 3x^2 + 2x - 1
To find the stationary points, set the derivative equal to zero and solve for x:
0 = 3x^2 + 2x - 1
Now we can solve this quadratic equation. We can factor it or use the quadratic formula:
x = (-2 ± √(2^2 - 4 * 3 * -1)) / (2 * 3)
x = (-2 ± √(4 + 12)) / 6
x = (-2 ± √16) / 6
x = (-2 ± 4) / 6
So, we have two possible values for x: x = (-2 + 4) / 6 = 2/6 = 1/3 and x = (-2 - 4) / 6 = -6/6 = -1.
(ii) To determine the nature of the stationary points, we need to evaluate the second derivative at these points. The second derivative is the derivative of the first derivative:
d^2y/dx^2 = d/dx (3x^2 + 2x - 1)
= 6x + 2
Now let's substitute the values of x we found earlier into the second derivative:
For x = 1/3:
d^2y/dx^2 = 6(1/3) + 2 = 2 + 2 = 4
For x = -1:
d^2y/dx^2 = 6(-1) + 2 = -6 + 2 = -4
Now we can analyze the results:
- If the second derivative is positive at a point, it implies a minimum point.
- If the second derivative is negative at a point, it implies a maximum point.
So, at x = 1/3, the stationary point has a positive second derivative (4), meaning it is a minimum point.
And at x = -1, the stationary point has a negative second derivative (-4), meaning it is a maximum point.
(b) (i) To find the equation of the tangent at x = 1, we need to find the gradient (slope) of the curve at that point. We can do this by evaluating the derivative at x = 1:
For x = 1:
dy/dx = 3(1^2) + 2(1) - 1 = 3 + 2 - 1 = 4
Therefore, the gradient (slope) of the tangent line is 4. Additionally, since the tangent to a curve at a given point has the same gradient as the curve at that point, the equation of the tangent line can be written as:
y - y1 = m(x - x1)
y - y1 = 4(x - 1)
y - y1 = 4x - 4
y = 4x - 4 + y1
We don't have a specific y-coordinate, y1, yet because we haven't evaluated the y-coordinate of the curve at x = 1. So, we will need that value to complete the equation.
(ii) To find the equation of the normal to the curve at x = 1, we need to find the gradient (slope) perpendicular to the tangent line. The perpendicular gradient is the negative reciprocal of the tangent's gradient.
So, the slope of the normal line will be -1/4. Using the same concept as above, we can write the equation of the normal line as:
y - y2 = (m_perp)(x - x2)
y - y2 = (-1/4)(x - 1)
y - y2 = -1/4x + 1/4
y = -1/4x + 1/4 + y2
As in the previous part, we don't have a specific y-coordinate, y2, yet because we haven't evaluated the y-coordinate of the curve at x = 1.