Critical Points

Definition

A critical point of the function \(f\) is any number \(c\) in the domain of \(f\) such that

\[f'(c)=0 \quad \text{or} \quad \text{$f'(c)$ does not exist.}\]

Critical points of \(f\) correspond to possible locations of relative extrema.

Example 1

Find all critical points of \(f(x)=2x^3-15x^2+36x+20\).

Step 1: Compute \(f'(x)\).

\[f'(x) = 6x^2-30x+36\]

Step 2: Find \(x\) such that \(f'(x) = 0\).

\[\begin{align*} 6x^2-30x+36 &= 6(x^2-5x+6)\\ &= 6(x-2)(x-3)\\ &= 0 \end{align*}\]

when \(x=2\) or when \(x=3\).

Step 3: Find \(x\) such that \(f'(x)\) does not exist.

Since \(f'(x)\) is polynomial, it exists for all real numbers.

Step 4: Verify that the values found in Steps 2 and 3 are in the domain of \(f\).

The domain of \(f(x)\) is all real numbers. Therefore, since both values are in the domain of \(f\), \(x=2\) and \(x=3\) are critical points of \(f\).

Example 2

Find all critical points of \(f(x)=\dfrac{1}{x^2-1}\).

Step 1: Compute \(f'(x)\).

\[\begin{align*} f'(x) &= \frac{d}{dx}(x^2-1)^{-1}\\ &= -(x^2-1)^{-2}\cdot 2x \\ &= -\frac{2x}{(x^2-1)^2} \end{align*}\]

Step 2: Find \(x\) such that \(f'(x) = 0\).

\(f'(x) = 0\) when \(2x = 0\), which occurs when \(x=0\).

Step 3: Find \(x\) such that \(f'(x)\) does not exist.

\(f'(x)\) does not exist when \((x^2-1)^2 = 0\), which occurs when \(x=1\) and when \(x=-1\).

Step 4: Verify that the values found in Steps 2 and 3 are in the domain of \(f\).

The domain of \(f(x)\) is all real numbers except \(x=1\) and \(x=-1\). Therefore, \(x=0\) is a critical point, but \(x=1\) and \(x=-1\) are not critical points since they are not in the domain of \(f\).