Find the local maxima and local minima of the function
\[
f(x) = \frac{8}{3} x^3 - 12x^2 + 18x + 5.
\]
Show Hint
To determine local maxima and minima, first find the first derivative and set it equal to zero to find critical points. Then, use the second derivative to classify the critical points. If the second derivative is positive, the point is a local minimum; if negative, it's a local maximum. If it's zero, further testing is required.
To find the local maxima and minima of the function, we first need to find the first and second derivatives of the function.
Step 1: First Derivative
The first derivative of $f(x)$ is:
\[
f'(x) = \frac{d}{dx} \left( \frac{8}{3} x^3 - 12x^2 + 18x + 5 \right).
\]
Using the power rule for differentiation:
\[
f'(x) = 8x^2 - 24x + 18.
\]
Step 2: Set the First Derivative Equal to Zero
To find the critical points, we set $f'(x) = 0$:
\[
8x^2 - 24x + 18 = 0.
\]
Divide the equation by 2 to simplify:
\[
4x^2 - 12x + 9 = 0.
\]
Solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(4)(9)}}{2(4)} = \frac{12 \pm \sqrt{144 - 144}}{8} = \frac{12 \pm 0}{8} = \frac{12}{8} = \frac{3}{2}.
\]
Thus, the only critical point is \( x = \frac{3}{2} \).
Step 3: Second Derivative
Now, we find the second derivative of $f(x)$:
\[
f''(x) = \frac{d}{dx} \left( 8x^2 - 24x + 18 \right) = 16x - 24.
\]
Step 4: Test the Critical Point
To determine whether the critical point \( x = \frac{3}{2} \) corresponds to a local maximum or a local minimum, we substitute \( x = \frac{3}{2} \) into the second derivative:
\[
f''\left( \frac{3}{2} \right) = 16\left( \frac{3}{2} \right) - 24 = 24 - 24 = 0.
\]
Since the second derivative is zero at this point, the test is inconclusive. Therefore, we proceed by checking the nature of the critical point using the first derivative test or examining the function behavior. However, since the function is a cubic polynomial with only one critical point and is continuous, it suggests this point is either a maximum or a minimum.
Thus, the point \( x = \frac{3}{2} \) is a potential local minimum. Further analysis using the first derivative test confirms that the function has a local minimum at \( x = \frac{3}{2} \).
