We can also define calculus critical value, as if we are having a function f(x) and we say that x = a is the critical point, if f (a) exists. Then, the following can be true:

f’(a) = 0 or f’(a) doesn't exists.

We can find the critical point, if we are having the knowledge of differentiation of all the functions. If we are asked to find the critical point of a polynomial function, then it is the simplest task in the case of critical point. This is because, by the degree of polynomial, we can determine how many critical point we will get.

If the degree is one, then critical point will be one and if degree is two, then critical point will be two and so on. But, in the case of trigonometric function, we can't determine the number of critical point.

Now, we will see one example of polynomial function for finding critical points.

**Example:**

Find the calculus critical value for the function given below:

f(x) = x

^{3}+ 2x

^{2}- 4x

**Solution:**

For finding the critical point, we have to first calculate the derivative of the function as shown below:

f’(x) = 3x

^{2}+ 4x - 4 (The derivative of x

^{3}is 3x

^{2}, the derivative of 2x

^{2}is 4x and derivative of 4x is 4).

Now, we will put that expression equals to zero for finding the critical point.

3x

^{2}- 4x - 4 = 0

For finding the critical point, we need to find the value of ‘x’ by factorizing the given equation,

3x

^{2}+ 6x - 2x - 4 = 0

3x(x + 2) - 2(x + 2) = 0,

(3x - 2)(x + 2) = 0

So, the required factors are x = $\frac{2}{3}$ and -2. So, critical points are $\frac{2}{3}$ and -2.