What are the critical numbers of the function:
f(x)=2x^3-3x^2-36x+14
Thanks
Nothing nice going on here
I tried, x = ±1, ±2, ±7, ±7/2, then gave up
ran it through Wolfram and got this
http://www.wolframalpha.com/input/?i=2x%5E3-3x%5E2-36x%2B14%3D0
To find the critical numbers of a function, we need to find the values of x where the derivative of the function is equal to zero or undefined.
Given the function:
f(x) = 2x^3 - 3x^2 - 36x + 14
To find the derivative, you can use the power rule and the sum or difference rule for differentiation. Differentiating each term of the function, we get:
f'(x) = 6x^2 - 6x - 36
Now, set the derivative equal to zero and solve for x:
6x^2 - 6x - 36 = 0
Factor the equation if possible or use the quadratic formula to solve for x. In this case, the equation can be factored:
2(3x^2 - 3x - 18) = 0
2(3x + 6)(x - 3) = 0
Setting each factor equal to zero yields two possible solutions:
3x + 6 = 0 or x - 3 = 0
Solving the first equation:
3x + 6 = 0
3x = -6
x = -2
Solving the second equation:
x - 3 = 0
x = 3
Therefore, the critical numbers of the given function f(x) = 2x^3 - 3x^2 - 36x + 14 are x = -2 and x = 3.