Step 1: Understanding the Concept:
A critical point of a function \( f(x) \) is defined as a point in its domain where the first derivative \( f'(x) \) is either zero or does not exist.
Since \( f(x) \) is a polynomial function, it is continuous and differentiable over the entire set of real numbers.
Therefore, we only need to find the values of \( x \) for which the derivative is exactly zero.
Identifying the number of critical points is essential for determining the intervals of increase or decrease and the local extrema of the function.
Step 2: Key Formula or Approach:
We will use the power rule of differentiation to find \( f'(x) \).
The power rule states that if \( y = x^{n} \), then \( \frac{dy}{dx} = n x^{n-1} \).
After finding the derivative, we will equate it to zero and solve the resulting polynomial equation for \( x \).
Step 3: Detailed Explanation:
The given function is:
\[ f(x) = x^{5} - 5x^{3} + 5x^{2} - 1 \]
Differentiating \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^{5}) - 5\frac{d}{dx}(x^{3}) + 5\frac{d}{dx}(x^{2}) - \frac{d}{dx}(1) \]
\[ f'(x) = 5x^{4} - 15x^{2} + 10x \]
To find the critical points, set \( f'(x) = 0 \):
\[ 5x^{4} - 15x^{2} + 10x = 0 \]
Divide the entire equation by 5 to simplify:
\[ x^{4} - 3x^{2} + 2x = 0 \]
Factor out the common term \( x \):
\[ x(x^{3} - 3x + 2) = 0 \]
This gives us our first critical point at \( x = 0 \).
Now, we must solve the cubic equation: \( x^{3} - 3x + 2 = 0 \).
Using the Rational Root Theorem, we test \( x = 1 \):
\[ (1)^{3} - 3(1) + 2 = 1 - 3 + 2 = 0 \]
Since \( x = 1 \) is a root, \( (x - 1) \) is a factor.
By synthetic division or polynomial long division:
\[ x^{3} - 3x + 2 = (x - 1)(x^{2} + x - 2) \]
Factoring the quadratic part \( x^{2} + x - 2 \):
\[ x^{2} + 2x - x - 2 = x(x + 2) - 1(x + 2) = (x - 1)(x + 2) \]
The fully factored derivative is:
\[ f'(x) = 5x(x - 1)(x - 1)(x + 2) = 5x(x - 1)^{2}(x + 2) \]
Setting each factor to zero, the roots are \( x = 0, x = 1, \) and \( x = -2 \).
Even though \( x = 1 \) is a repeated root, it represents a single unique point in the domain where the tangent is horizontal.
Thus, there are 3 distinct critical points: \( \{ -2, 0, 1 \} \).
Step 4: Final Answer:
The number of critical points is 3.