f(X)= (x^2)/(x^2 - 1) .find the symmetry , intercepts, asymptotes, interval of increase and decrease, local extreme values, concavity, point of deflection and draw the graph of the function.kindly show the working.
review your basic rules in this stuff.
f = x^2/(x^2-1)
f' = -2x/(x^2-1)^2
f" = 2(3x^2+1)/(x^2-1)^3
I expect you can find intercepts ok.
Since f(x) is even, it's symmetric about the y-axis.
vertical asymptotes where x^2-1 = 0
Since f = 1 + 1/(2(x^2-1))
horizontal asymptote is at y=1
f is increasing where f' > 0
f is concave up where f" > 0
You can check your calculations by viewing the graph at
http://www.wolframalpha.com/input/?i=+x^2%2F%28x^2-1%29
To find the symmetry of the function f(x) = (x^2)/(x^2 - 1), we can check if it is an even function (symmetric about the y-axis) or an odd function (symmetric about the origin).
For an even function, we check if f(x) = f(-x) for all x in the domain.
For an odd function, we check if f(x) = -f(-x) for all x in the domain.
Let's first find the intercepts of the function f(x).
To find the x-intercepts, we set f(x) = 0 and solve for x:
(x^2) / (x^2 - 1) = 0
Since the numerator is 0 when x = 0, the x-intercept is x = 0.
To find the y-intercept, we substitute x = 0 into the function:
f(0) = (0^2) / (0^2 - 1) = 0 / (-1) = 0
Therefore, the y-intercept is y = 0.
Next, let's find the vertical asymptotes of the function.
Vertical asymptotes occur when the denominator of the function approaches zero.
So, we set the denominator (x^2 - 1) equal to zero and solve for x:
x^2 - 1 = 0
(x - 1)(x + 1) = 0
x - 1 = 0 or x + 1 = 0
x = 1 or x = -1
Therefore, the vertical asymptotes are x = 1 and x = -1.
Now, let's determine the intervals of increase and decrease.
We can find the critical points by finding the values of x where the derivative of f(x) is equal to zero or undefined.
Let's find the derivative of f(x) to determine the intervals of increase and decrease:
f'(x) = [2x(x^2 - 1) - x^2(2x)] / (x^2 - 1)^2
Simplifying further:
f'(x) = [2x^3 - 2x - 2x^3] / (x^2 - 1)^2
f'(x) = [-2x] / (x^2 - 1)^2
f'(x) = -2x / (x^2 - 1)^2
To find the critical points, we set the derivative equal to zero:
-2x / (x^2 - 1)^2 = 0
Since the numerator is zero when x = 0, the critical point is x = 0.
Let's create a number line and test the intervals.
Test the interval (-∞, -1):
Choose x = -2:
f'(-2) = -2(-2) / ((-2)^2 - 1)^2 = 4 / 9 > 0
The function is increasing on the interval (-∞, -1).
Test the interval (-1, 0):
Choose x = -0.5:
f'(-0.5) = -2(-0.5) / ((-0.5)^2 - 1)^2 = 1.5 < 0
The function is decreasing on the interval (-1, 0).
Test the interval (0, 1):
Choose x = 0.5:
f'(0.5) = -2(0.5) / ((0.5)^2 - 1)^2 = -1.5 < 0
The function is decreasing on the interval (0, 1).
Test the interval (1, ∞):
Choose x = 2:
f'(2) = -2(2) / ((2)^2 - 1)^2 = 4 / 9 > 0
The function is increasing on the interval (1, ∞).
Now let's determine the concavity and points of inflection.
To find the second derivative of f(x), differentiate f'(x):
f''(x) = [d/dx (-2x)] / (x^2 - 1)^2
f''(x) = -2 / (x^2 - 1)^2
Since the second derivative is a constant, it does not equal zero or undefined. Therefore, there are no points of inflection in the domain.
Finally, let's draw the graph of the function f(x).
From the information gathered above:
- The function has a y-intercept of 0.
- It has a vertical asymptote at x = 1 and x = -1.
- The function is increasing on the intervals (-∞, -1) and (1, ∞).
- The function is decreasing on the intervals (-1, 0) and (0, 1).
- There are no local extreme values.
- The concavity is negative everywhere.
Since creating an accurate graph using only text is challenging, it would be best to input the function into a graphing software or calculator to visualize its shape.