Sketch the graphs of the functions - Calculus question

Click For Summary
The discussion focuses on sketching the graph of the polynomial function y = x^4 - 4x^3 - 20x^2 + 150, analyzing its critical points, intervals of increase and decrease, concavity, and local extrema. The function has no asymptotes or symmetries, with critical points identified at x = -2, 0, and 5, leading to a local minimum at (-2, 118) and a local maximum at (0, 150). The second derivative indicates concavity changes, revealing intervals of concave up and concave down behavior. The y-intercept is confirmed at 150, and the zeros of the function are approximately 2.5139 and 6.5252. Overall, the calculations and findings are validated by peer feedback, suggesting confidence in the results.
Mary4ever
Messages
11
Reaction score
0

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
 
Physics news on Phys.org
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K