a. Find the open intervals on which the function is...

$\begin{equation} \begin{array}{l}{\text { a. Find the open intervals on which the function is increasing and }} \\ {\text { decreasing. }} \\ {\text { b. Identify the function's local and absolute extreme values, if }} \\ {\text { any, saying where they occur. }}\end{array} \end{equation}$ $$\begin{equation} H(t)=\frac{3}{2} t^{4}-t^{6} \end{equation}$$

$\begin{equation}
\begin{array}{l}{\text { a. Find the open intervals on which the function is increasing and }} \\ {\text { decreasing. }} \\ {\text { b. Identify the function's local and absolute extreme values, if }} \\ {\text { any, saying where they occur. }}\end{array}
\end{equation}$
$$\begin{equation}
H(t)=\frac{3}{2} t^{4}-t^{6}
\end{equation}$$

Key Concepts

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate of change of the function relative to its independent variable. Calculating the derivative is essential when analyzing the behavior of functions, including identifying where they increase or decrease and locating potential extreme values.

Critical Points

Critical points are the values of the independent variable where the derivative is zero or undefined. These points are important because they are the potential candidates where the function might have local maxima or minima, thus playing a central role in optimization and curve sketching.

Monotonicity Analysis

Monotonicity analysis involves determining the intervals where a function is increasing or decreasing by examining the sign of its derivative. If the derivative is positive over an interval, the function is increasing on that interval; if negative, the function is decreasing. This analysis is key for understanding the overall behavior of the function.

Local and Absolute Extreme Values

Local extremes are points where the function reaches a maximum or minimum within a small neighboring interval, while absolute extremes are the highest or lowest values the function attains over its entire domain. Identifying these extreme values along with their corresponding intervals is an important step in evaluating a function's global behavior and is particularly useful in optimization problems.

Transcript

00:01 Suppose h of t equals 3 halves times t raised to the 4th power minus t to the 6th.

00:06 For the first part, we want to find the open intervals in which the function is increasing and decreasing.

00:12 So to do this, we first want to get h prime of t, and this is just 3 halves times 4t raised to the 3rd power minus 6t raised to the 5th power.

00:24 That's equal to 6 t raised to the third power minus 6 t raised to the fifth power.

00:32 And then we want to set h prime of t equal to zero, solve for t, to get the critical numbers of the function.

00:40 That means you have 6 t raised to the third power minus 6 t raised to the fifth power equal to zero.

00:45 We factor out 6t raise to the third power.

00:48 That's going to give us 1 minus t squared equal to 0.

00:53 That means t is zero or t equals plus or minus one and then you want to partition our domain which in this case is all real numbers from negative infinity to infinity using the critical numbers zero negative one and one so let's say this is our number line then you'll partition this you have negative one zero and then one so the possible interval will be from negative infinity to negative 1, negative 1 to 0, 0 to 1, and then 1 to infinity.

01:32 And then you want to do first derivative test to get the sign and the conclusion for the behavior of the function.

01:42 And then to get the sign, we need the test values for each interval.

01:47 So let's say we use negative 2 here, negative 0 .5, positive 0 .5, and then 2.

01:55 2 here.

01:56 When t is equal to negative 2, if we are to use the factored form of h prime, which is this, this 6t raise of the third power is going to be negative, and 1 minus t squared is also negative, so that makes a positive h prime.

02:14 If it's negative 0 .5, 6t cube is still negative, but 1 minus t squared becomes positive, so h prime is now negative.

02:24 If it's positive, 0 .5 ,000, 0 .5, the cubic factor is positive, but 1 minus t squared is still positive...

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