Sketch the graphs of the functions - Calculus question

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Homework Help Overview

The discussion revolves around sketching the graph of a polynomial function of degree 4, specifically the function y = x^4 - 4x^3 - 20x^2 + 150. Participants are exploring various characteristics of the function, including intervals of increase and decrease, concavity, critical points, and intercepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to analyze the function by calculating its first and second derivatives to identify critical points and intervals of concavity. Some participants question the accuracy of the calculations and seek verification of the results.

Discussion Status

Participants are engaged in verifying the original poster's findings. Some express confidence in the calculations, while others request a double-check to ensure correctness. There is no explicit consensus, but the discussion is focused on confirming the details of the analysis.

Contextual Notes

The original poster expresses uncertainty about the correctness of their calculations and seeks assistance in confirming their findings. There is mention of using external tools for verification, indicating a reliance on technology for support.

Mary4ever
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Homework Statement


Sketch the graphs of the functions. Indicate intervals on which the function is increasing, decreasing, concave up, or concave down; indicate relative maximum points, relative minimum points, points of inflection, horizontal asympototes, vertical asymptotes, symmetry, and those intercepts that can be obtained conveniently:

Homework Equations


y=x^4 - 4x^3 - 20x^2 +150


The Attempt at a Solution


This is the solution I have:
This is a polynomial of degree 4, so there are no asymptotes.
The function has no symmetries.
dy/dx = 4x^3 - 12x^2 - 40x
d^2y/dx^2 = 12x^2 - 24x - 40

dy/dx = 4(x^3 - 3x^2 - 10x) = 4x(x-5)(x+2)

There are 0s at x = -2, 0, and 5
Thus, there are three critical points for a 4th degree polynomial.
As the function goes to infinity as x goes to plus or minus infinity, we know that
the function decreases on (-infinity, -2), increases on (-2, 0), decreases on (0,5), and increases on (5,infinity)
f(-2) = (-2)^4 -4*(-2)^3 - 20*(-2)^2 + 150 = 16 + 32 - 80 + 150 = 118
There is a local minimum at (-2, 118)
f(0) = 150
There is a local maximum at (0, 150)
f(5) = (5)^4 -4*(5)^3 - 20*(5)^2 + 150 = 625 - 500 - 500 + 150 = -225
There is a local minimum at (5, -225)

The y-intercept is 150

The 0s are approximately 2.5139 and 6.5252

Finally, as the second derivative is 12x^2 - 24x - 40 = 4(3x^2 - 6x - 10), the inflection points are
1 plus or minus sqrt(39)/3 is approximately -1.08166599946613 and 3.08166599946613

As the second derivative is quadratic with a positive leading coefficient, we then know that
the function is concave up on (-infinity, 1 - sqrt(39)/3) or (-infinity, -1.08166599946613 ) and
(1 + sqrt(39)/3,infinity) or (3.08166599946613, infinity) and
concave down on (1 - sqrt(39)/3,1 + sqrt(39)/3) or (-1.08166599946613,3.08166599946613)

But I am not sure if it is correct. Please help
 
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Welcome to PF!

Hi Mary4ever! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)

Yes, that all looks fine. :smile:

(I haven't checked the y coordinate calculations)

What is worrying you about that? :confused:
 
Could you please double-check it because I need to make sure everything is correct? Thank you!
 
This all looks good, based on inspection of the graph and a little help from WolframAlpha.
 

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