# How do you find the critical points for #y = 2x^3 + 3x^2 - 12x + 4#?

You evaluate the first derivative and find its zeroes:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the critical points of the function y = 2x^3 + 3x^2 - 12x + 4, you first need to find its derivative, y'. Then, set y' equal to zero and solve for x. The values of x obtained from this equation will be the critical points.

y' = 6x^2 + 6x - 12

Set y' equal to zero: 6x^2 + 6x - 12 = 0

Now, solve for x using methods like factoring, completing the square, or the quadratic formula.

Using the quadratic formula: x = [-b ± √(b^2 - 4ac)] / (2a)

Where a = 6, b = 6, and c = -12.

Plugging in these values: x = [-6 ± √(6^2 - 4 * 6 * (-12))] / (2 * 6)

x = [-6 ± √(36 + 288)] / 12

x = [-6 ± √324] / 12

x = [-6 ± 18] / 12

So, the critical points are: x = (-6 + 18) / 12 = 12 / 12 = 1 x = (-6 - 18) / 12 = -24 / 12 = -2

Therefore, the critical points for the function y = 2x^3 + 3x^2 - 12x + 4 are x = 1 and x = -2.

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the intervals of increasing and decreasing given #y=x^3-11x^2+39x-47#?
- How do use the first derivative test to determine the local extrema #3-x^2#?
- Is #f(x)=(-x^3+x^2-5x+6)/(x-2)# increasing or decreasing at #x=0#?
- What are the values and types of the critical points, if any, of #f(x,y) = x(2-x-y)#?
- What are the critical points of #f(x,y) = x^3 + y^3 - xy#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7