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 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#
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 \(f\).