How do I find the minimum or maximum of this y = 2x³ + 3x² - 12x? It makes 2 curves.
y = 2x^3 + 3x^2 - 12x
y' = 6x^2 + 6x - 12 = 6(x+2)(x-1)
y'=0 when x=-2 or 1
Knowing what you do about cubic curves, you can see that
the local max occurs at x = -2, and
the local min is at x = 1
y(-2) = 20
y(1) = -7
To find the minimum or maximum of a function, we need to determine the critical points where the derivative of the function is equal to zero.
To find the derivative of the function y = 2x³ + 3x² - 12x, we can use the power rule. The derivative of x^n is nx^(n-1).
Let's differentiate y = 2x³ + 3x² - 12x step by step:
Step 1: Take the derivative of each term separately:
dy/dx = d/dx (2x³) + d/dx (3x²) - d/dx (12x)
Step 2: Apply the power rule to each term:
dy/dx = 3 * 2x^(3-1) + 2 * 3x^(2-1) - 12 * 1x^(1-1)
Step 3: Simplify the exponents and constants:
dy/dx = 6x² + 6x - 12
Now, we can find the critical points by setting the derivative equal to zero and solving for x:
6x² + 6x - 12 = 0
Step 4: Factor or use the quadratic formula to solve for x:
We can factor out a common factor of 6:
6(x² + x - 2) = 0
Now, we can factor the quadratic equation:
6(x - 1)(x + 2) = 0
Set each factor equal to zero:
x - 1 = 0 or x + 2 = 0
Solving for x, we find:
x = 1 or x = -2
These are the critical points where the derivative is equal to zero.
To determine if these are minimum or maximum points, we need to find the second derivative.
The second derivative is the derivative of the first derivative. Let's differentiate dy/dx = 6x² + 6x - 12:
d²y/dx² = d/dx (6x² + 6x - 12)
Using the power rule again:
d²y/dx² = 2 * 6x^(2-1) + 1 * 6x^(1-1)
Simplifying this further:
d²y/dx² = 12x + 6
Now, we can substitute the critical points we found earlier (x = 1 and x = -2) into the second derivative:
For x = 1:
d²y/dx² = 12(1) + 6 = 12 + 6 = 18
For x = -2:
d²y/dx² = 12(-2) + 6 = -24 + 6 = -18
The second derivative at x = 1 is positive (18) and at x = -2 is negative (-18).
If the second derivative is positive, it indicates a minimum point, and if the second derivative is negative, it indicates a maximum point.
Therefore, at x = 1, we have a minimum point, and at x = -2, we have a maximum point.
To find the corresponding y-values, substitute the critical points into the original function:
For x = 1:
y = 2(1)³ + 3(1)² - 12(1) = 2 + 3 - 12 = -7
For x = -2:
y = 2(-2)³ + 3(-2)² - 12(-2) = -16 + 12 + 24 = 20
So, the minimum point is (1, -7) and the maximum point is (-2, 20).
To find the minimum or maximum of a function, we need to first determine the critical points. Critical points are the values of x where the derivative of the function is equal to zero or undefined.
Let's find the derivative of the function y = 2x³ + 3x² - 12x. The derivative, denoted as dy/dx or f'(x), represents the rate of change of the function at any given point.
To find the derivative, we apply the power rule and sum rule of differentiation. The power rule states that when we differentiate x^n with respect to x, where n is a constant, we get n*x^(n-1).
So, for the given function y = 2x³ + 3x² - 12x:
dy/dx = d/dx(2x³) + d/dx(3x²) + d/dx(-12x)
= 6x² + 6x - 12
Now, let's find the critical points by setting the derivative equal to zero and solving for x:
6x² + 6x - 12 = 0
To solve this quadratic equation, we can factor it if possible or use the quadratic formula. In this case, factoring is not straightforward, so we'll use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
For our equation, a = 6, b = 6, and c = -12. Plugging these values into the quadratic formula:
x = (-6 ± √(6² - 4(6)(-12))) / (2(6))
x = (-6 ± √(36 + 288)) / 12
x = (-6 ± √324) / 12
x = (-6 ± 18) / 12
This gives us two possible values for x: x = (-6 + 18) / 12 = 12/12 = 1, and x = (-6 - 18) / 12 = -24/12 = -2.
Therefore, the critical points are x = 1 and x = -2.
To determine if these critical points correspond to minimum or maximum points, we can use the second derivative test. The second derivative, denoted as d²y/dx² or f''(x), represents the concavity of the function.
To find the second derivative, we differentiate the first derivative with respect to x:
d²y/dx² = d/dx(6x² + 6x - 12)
= 12x + 6
Now, plug in the critical points x = 1 and x = -2 into the second derivative equation:
For x = 1: d²y/dx² = 12(1) + 6 = 12 + 6 = 18
For x = -2: d²y/dx² = 12(-2) + 6 = -24 + 6 = -18
Since the second derivative is positive (18) at x = 1, the function has a minimum point at x = 1. And since the second derivative is negative (-18) at x = -2, the function has a maximum point at x = -2.
To find the corresponding y-values of these extreme points, substitute the x-values into the original function:
For x = 1: y = 2(1)³ + 3(1)² - 12(1) = 2 + 3 - 12 = -7
For x = -2: y = 2(-2)³ + 3(-2)² - 12(-2) = -16 + 12 + 24 = 20
Therefore, the minimum point is (1, -7) and the maximum point is (-2, 20).