Critical Points

Definition

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 versus Relative Extrema

Critical points of \(f\) correspond to possible locations of relative extrema. In other words, not all critical points correspond to a relative extrema. However, every relative extrema must appear at a critical point.

Example 1

Finding critical points

Find the critical points of \(f(x) = \sqrt[3]{x^2-1}\).

Step 1:   Compute   \(f'(x)\).

\[\begin{align*} f'(x) &= \frac{d}{dx} (x^2-1)^{1/3} \\ &= \frac{1}{3}(x^2 - 1)^{-2/3} (2x) \\ &= \frac{2x}{3(x^2 - 1)^{2/3}} \end{align*}\]

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

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

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

\(f'(x)\) does not exist when \(x^2-1 = 0\), 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 \(f\) is all real numbers. Therefore, \(x=-1\), \(x=0\), and \(x=1\) are all critical points of \(f\).

Example 2

Finding critical points

Find the critical points of \(f(x) = x^3 +3x^2 - 24x + 1\).

Step 1:   Compute   \(f'(x)\).

\[\begin{align*} f'(x) &= 3x^2 + 6x - 24 \end{align*}\]

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

\[\begin{align*} f'(x) &= 3(x^2 + 2x - 8) \\ &= 3(x+4)(x-2) \end{align*}\]

which equals zero when \(x=-4\) and \(x=2\).

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

Since \(f'(x)\) is a 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 \(f\) is all real numbers. Therefore, \(x=-4\) and \(x=2\) are both critical points of \(f\).

Example 3

Finding critical points

Find the critical points of \(f(x) = \dfrac{x}{x^2-4}\).

Step 1:   Compute   \(f'(x)\).

\[\begin{align*} f'(x) &= \frac{d}{dx} \dfrac{x}{x^2-4} \\ &= \frac{1(x^2-1) - x(2x)}{(x^2-4)^2} && \text{quotient rule}\\ &= \frac{x^2 - 1 - 2x^2}{(x^2-4)^2} && \text{simplify numerator}\\ &= \frac{- 1 - x^2}{(x^2-4)^2} \\ &= -\frac{1 + x^2}{(x^2-4)^2} \end{align*}\]

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

\(f'(x)=0\) when \(1+x^2 = 0\). However, there are no values of \(x\) such that \(1 + x^2 = 0\). Therefore, \(f'(x)\) is never equal to zero.

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

\(f'(x)\) does not exist when \(x^2-4 = 0\), which occurs when \(x=2\) and when \(x=-2\).

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

The domain of \(f\) is all real numbers except \(x = \pm 2\), which means that the values found in Step 3 are not critical points since they are not in the domain of \(f\). And since there were no other values found in Step 2, \(f\) does not have any critical points.