Question

for the following exercise, use a graphing utility to estimate the local extrema of the function and to estimate the intervals on which the function is increasing and decreasing.
$f(t) = 3t^{\frac{2}{3}} - t$
furthest left local extrema: (, )
furthest right local extrema: (, )
increasing on the interval:
decreasing on the interval:
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Explanation:

Step1: Find derivative of $f(t)$

$f'(t) = 3\times\frac{2}{3}t^{-\frac{1}{3}} - 1 = \frac{2}{t^{\frac{1}{3}}} - 1$

Step2: Find critical points

Set $f'(t)=0$:
$\frac{2}{t^{\frac{1}{3}}} - 1 = 0$
$\frac{2}{t^{\frac{1}{3}}} = 1$
$t^{\frac{1}{3}} = 2$
$t = 8$
Also, $f'(t)$ is undefined at $t=0$, so critical points are $t=0$ and $t=8$.

Step3: Evaluate $f(t)$ at critical points

At $t=0$: $f(0)=3(0)^{\frac{2}{3}} - 0 = 0$
At $t=8$: $f(8)=3(8)^{\frac{2}{3}} - 8 = 3\times4 - 8 = 12-8=4$

Step4: Test intervals for increase/decrease

• For $t<0$: Choose $t=-1$, $f'(-1)=\frac{2}{(-1)^{\frac{1}{3}}}-1=-2-1=-3<0$, function decreasing.
• For $00$, function increasing.
• For $t>8$: Choose $t=27$, $f'(27)=\frac{2}{27^{\frac{1}{3}}}-1=\frac{2}{3}-1=-\frac{1}{3}<0$, function decreasing.

Step5: Classify local extrema

At $t=0$ (leftmost critical point): function changes from decreasing to increasing, so local minimum.
At $t=8$ (rightmost critical point): function changes from increasing to decreasing, so local maximum.

Answer:

Furthest left local extrema: $(0, 0)$
Furthest right local extrema: $(8, 4)$
Increasing on the interval: $(0, 8)$
Decreasing on the interval: $(-\infty, 0) \cup (8, \infty)$