Math Problem Statement

Describe the left-hand and right-hand behavior of the graph of the function, then sketch its graph. g(t) = -(2/9)t^4 + 3t

Solution

The function given is:

g(t)=29t4+3tg(t) = -\frac{2}{9}t^4 + 3t

Left-hand and right-hand behavior:

  1. As tt \to -\infty (left-hand behavior):

    • The highest degree term dominates the behavior of the function as tt \to -\infty. The term 29t4-\frac{2}{9}t^4 grows very large negatively, since t4t^4 will grow large and positive, but the negative sign in front makes the whole term negative.
    • The linear term 3t3t contributes a smaller effect because t4t^4 dominates. Therefore, as tt \to -\infty, g(t)g(t) \to -\infty.
  2. As tt \to \infty (right-hand behavior):

    • Similarly, for tt \to \infty, the term 29t4-\frac{2}{9}t^4 dominates. Since t4t^4 becomes very large and the negative coefficient makes it negative, the function will tend to -\infty as well.
    • The linear term 3t3t again has a smaller influence in comparison to the dominant t4t^4-term. Thus, as tt \to \infty, g(t)g(t) \to -\infty.

Critical points:

To better understand the behavior and find any critical points, we differentiate g(t)g(t) with respect to tt:

g(t)=89t3+3g'(t) = -\frac{8}{9}t^3 + 3

Setting g(t)=0g'(t) = 0 to find critical points:

89t3+3=0-\frac{8}{9}t^3 + 3 = 0 t3=278t^3 = \frac{27}{8} t=2783=32t = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}

So, t=32t = \frac{3}{2} is a critical point. We can analyze the second derivative to classify the critical point as a local maximum or minimum.

g(t)=249t2=83t2g''(t) = -\frac{24}{9}t^2 = -\frac{8}{3}t^2

At t=32t = \frac{3}{2}:

g(32)=83×(32)2=6g''\left( \frac{3}{2} \right) = -\frac{8}{3} \times \left( \frac{3}{2} \right)^2 = -6

Since g(t)<0g''(t) < 0, the function has a local maximum at t=32t = \frac{3}{2}.

Sketch of the graph:

  • The function has a local maximum at t=32t = \frac{3}{2}, and the value of g(t)g(t) at this point can be found by substituting t=32t = \frac{3}{2} into the original function.
  • The graph falls towards -\infty on both sides (left and right behavior), with the local maximum in between.

Let me know if you'd like me to generate the graph.


Do you want any details or clarifications on this?

Relative questions:

  1. What is the significance of critical points in understanding a function's graph?
  2. How does the degree of a polynomial affect its end behavior?
  3. What is the importance of the second derivative in determining concavity?
  4. Can a function have more than one local maximum or minimum?
  5. How do inflection points relate to the curvature of a graph?

Tip: When analyzing polynomials, the highest degree term typically dictates the end behavior of the graph.

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Math Problem Analysis

Mathematical Concepts

Polynomial Functions
End Behavior
Critical Points
Concavity

Formulas

Derivative: g'(t) = -(8/9)t^3 + 3
Second Derivative: g''(t) = -(8/3)t^2

Theorems

End Behavior Theorem
Critical Points from Derivatives
Concavity Test

Suitable Grade Level

Grades 10-12