For the function y=4-2x^2+1/6x^4 find the following:
Domain
x and y intercepts
Vertical asymptotes
Horizontal asymptotes
Symmetry
F'(x)
Critical numbers
Increasing f(x)
Decreasing f(x)
Extrema
F"(x)
Possible points of inflection
Concave up
Concave down
Points of inflection
I got some:
domain: (-infinity, infinity)
y-intercept: y=4
Vertical asymptotes: none
Horizontal asymptotes: none
Symmetry: y-axis
F'(x): 2/3x^3-4x
Critical numbers: x=0, x=sqrt6, x=-sqrt6
Extrema: relative maximum f(0)=4, relative minimums f(-sqrt2)=-2 and f(sqrt2)=-2
F"(x): 2x^2-4
Possible points of inflection: (sqrt2, 2/3) (-sqrt2, 2/3)
Concave up: (-infinity, -sqrt2) and (sqrt2, infinity)
Concave down: (-sqrt2, sqrt2)
To make it clearer, I can't figure out
x-intercept
Increasing f(x)
Decreasing f(x)
Points of inflection
I appreciate your time very much. Thank you.
taking your typing at face value the way you typed it
your domain is correct
as a matter of fact:
"I got some:
domain: (-infinity, infinity)
y-intercept: y=4
Vertical asymptotes: none
Horizontal asymptotes: none
Symmetry: y-axis
F'(x): 2/3x^3-4x
Critical numbers: x=0, x=sqrt6, x=-sqrt6
Extrema: relative maximum f(0)=4, relative minimums f(-sqrt2)=-2 and f(sqrt2)=-2
F"(x): 2x^2-4
Possible points of inflection: (sqrt2, 2/3) (-sqrt2, 2/3)
Concave up: (-infinity, -sqrt2) and (sqrt2, infinity)
Concave down: (-sqrt2, sqrt2) "
all looks good up to here
for x-intercept , we let y = 0
4-2x^2+1/6x^4 = 0
times 6
24 - 12x^2 + x^4 = 0
let's complete the square:
x^4 - 12x^2 + 36 = -24+36
(x^2 - 6)^2 = 12
x^2 - 6 = ± √12
x^2 = 6 ±√12
x = ±√(6 ±√12) , so we have 4 x-intercepts
pts of inflection, f '' (x) = 0
2x^2 - 4 = 0
x^2 = 2
x = ±√2
if x = √2 , y = 4-4+4/6 = 2/3
if x = -√2, y = 2/3
(you had that correct)
for increasing function:
f ' (x) > 0
(2/3)x^3 - 4x > 0
2x^3 - 12x > 0
x^3 - 6x > 0
x(x^2 - 6) > 0
x(x + √6)(x - √6) > 0
which is true for -√6 < x < 0 OR x > √6
it then follows that the function would be decreasing for
x < -√6 or 0 < x < √6
here is a picture of your graph:
http://www.wolframalpha.com/input/?i=plot+y%3D4-2x%5E2%2B1%2F6x%5E4
To find the x-intercepts, we need to set y equal to zero and solve for x. So, we have:
0 = 4 - 2x^2 + 1/6x^4
To solve this equation, we can factor it if possible. However, in this case, it isn't possible to factor it easily. So, we can use numerical methods such as graphing or using a calculator to find the x-intercepts. By graphing or using a suitable software, you can find the approximate values of the x-intercepts.
To determine where the function is increasing or decreasing, we need to examine the intervals of the function by determining where the first derivative, f'(x), is positive (increasing) or negative (decreasing).
Given the first derivative, f'(x) = 2/3x^3 - 4x, we need to find the critical numbers where the derivative is equal to zero or undefined.
Setting f'(x) equal to zero, we have:
0 = 2/3x^3 - 4x
To solve this equation, we can factor out x:
0 = x(2/3x^2 - 4)
Setting each factor equal to zero, we find two critical numbers:
x = 0, 2/3x^2 - 4 = 0
Simplifying the second equation, we get:
2/3x^2 - 4 = 0
2/3x^2 = 4
x^2 = (4 * 3) / 2
x^2 = 6
x = ±√6
Now, we can create an interval chart to determine where the function is increasing or decreasing:
Interval: (-∞, -√6) | (-√6, 0) | (0, √6) | (√6, +∞)
f'(x): negative | positive | negative | positive
f(x): decreasing | increasing | decreasing | increasing
Next, to identify the points of inflection, we need to find where the second derivative, f''(x), is equal to zero or undefined. The second derivative is given by:
f''(x) = 2x^2 - 4
Setting f''(x) equal to zero, we have:
0 = 2x^2 - 4
2x^2 = 4
x^2 = 2
x = ±√2
So, the possible points of inflection are (±√2, f(±√2)), where f(±√2) can be found by substituting the values into the original function: y = 4 - 2x^2 + 1/6x^4.
For points of inflection, we need to determine the concavity of the function. By analyzing the signs of the second derivative, we can identify where the function is concave up or concave down.
Interval: (-∞, -√2) | (-√2, √2) | (√2, +∞)
f''(x): negative | positive | negative
f(x): concave down | concave up | concave down
I hope this explanation clarifies the x-intercepts, increasing f(x), decreasing f(x), and points of inflection for the given function. Let me know if you have any further questions!