(1 point) The derivative of $f(x)$ is given by $f^{\prime}(x)=\frac{(2-x)(x+3)}{(x-3)(x+2)}$. Find the critical points and local extrema of $f$, and the open intervals on which $f$ is increasing and decreasing. (Assume that $f(x)$ and $f^{\prime}(x)$ have the same domain.)
Critical points of $f(x)$ are
(If there is more than one critical point, enter a comma-separated list. If there are no criticahpoints, enter "none".)
$f(x)$ is increasing on
$f(x)$ is decreasing on
(Enter your answers using interval notation.)
$f(x)$ has local maxima at
$f(x)$ has local minima at
(If there is more than one extremum of a given type, enter a comma-separated list. If there are no extrema of a given type, enter "none".)
Answer
\(\boxed{\text{The function has local maxima at } x = -3, 2 \text{ and local minima at } x = -2, 3}\)
Step 1 ：Find the critical points of the function by setting the derivative equal to zero: \((2-x)(x+3) = 0\). This gives us two solutions: \(x = 2\) and \(x = -3\).
Step 2 ：Find where the derivative is undefined by setting the denominator equal to zero: \((x-3)(x+2) = 0\). This gives us two more solutions: \(x = 3\) and \(x = -2\).
Step 3 ：The critical points of the function are \(x = 2\), \(-3\), \(3\), \(-2\).
Step 4 ：Determine where the function is increasing and decreasing by testing the sign of the derivative on the intervals determined by the critical points. Choose test points -4, -2.5, 0, 2.5, and 4.
Step 5 ：For \(x = -4\), \(f'(x)\) is positive. For \(x = -2.5\), \(f'(x)\) is negative. For \(x = 0\), \(f'(x)\) is positive. For \(x = 2.5\), \(f'(x)\) is negative. For \(x = 4\), \(f'(x)\) is positive.
Step 6 ：The function is increasing on the intervals \((-\infty, -3)\) and \((-2, 2)\) and \((3, \infty)\), and decreasing on the intervals \((-3, -2)\) and \((2, 3)\).
Step 7 ：Determine the local extrema. The function changes from increasing to decreasing at \(x = -3\) and \(x = 2\), these are local maxima. The function changes from decreasing to increasing at \(x = -2\) and \(x = 3\), these are local minima.
Step 8 ：\(\boxed{\text{The function has local maxima at } x = -3, 2 \text{ and local minima at } x = -2, 3}\)
