Question

Исследуйте функцию: y=3-x^2/x+2 на монотонность и экстремумы

Answer 1

Y=(3-x²)/(x+2)
y`=(-2x*(x+2)-1*(3-x²))/(x+2)²=(-2x²-4x-3+x²)/(x+2)²=-(x²+4x+3)/(x+2)²=0
-(x²+4x+3)=0
x1+x2=-4 u x1*x2=3
x1=-3 U x2=-1
           _                  +                    _
---------------------------------------------------
 убыв            -3 возр          -1  убыв
                   min                max
убыв x∈(-∞;-3) U (1;∞)
возр x∈(-3;-1)                  

Answer 2

Function Analysis: y = (3 - x^2) / (x + 2)

To analyze the function y = (3 - x^2) / (x + 2) for monotonicity and extrema, we need to examine its derivative and second derivative.

Let's start by finding the derivative of the function.

The derivative of y with respect to x can be found using the quotient rule:

y' = [(d/dx)(3 - x^2)(x + 2) - (3 - x^2)(d/dx)(x + 2)] / (x + 2)^2

Simplifying this expression, we get:

y' = [(2x)(x + 2) - (3 - x^2)] / (x + 2)^2

Expanding and simplifying further:

y' = (2x^2 + 4x - 3 + x^2) / (x + 2)^2

Combining like terms:

y' = (3x^2 + 4x - 3) / (x + 2)^2

Now, let's find the second derivative of y.

The second derivative of y can be found by differentiating y' with respect to x:

y'' = [(d/dx)(3x^2 + 4x - 3)(x + 2)^2 - (3x^2 + 4x - 3)(d/dx)(x + 2)^2] / (x + 2)^4

Simplifying this expression, we get:

y'' = [(6x + 4)(x + 2)^2 - (3x^2 + 4x - 3)(2(x + 2))] / (x + 2)^4

Expanding and simplifying further:

y'' = (6x^3 + 24x^2 + 28x + 8 - 6x^3 - 8x^2 + 6x^2 + 8x - 6) / (x + 2)^4

Combining like terms:

y'' = (22x^2 + 36x + 2) / (x + 2)^4

Now that we have the first and second derivatives of y, we can analyze the function for monotonicity and extrema.

Monotonicity Analysis:

To determine the monotonicity of the function, we need to examine the sign of the first derivative, y'.

Let's find the critical points by setting y' equal to zero and solving for x:

(3x^2 + 4x - 3) / (x + 2)^2 = 0

Multiplying both sides by (x + 2)^2 to eliminate the denominator:

3x^2 + 4x - 3 = 0

Using the quadratic formula, we can solve for x:

x = (-4 ± √(4^2 - 4 * 3 * (-3))) / (2 * 3)

Simplifying this expression, we get:

x = (-4 ± √(16 + 36)) / 6

x = (-4 ± √52) / 6

Simplifying further:

x = (-2 ± √13) / 3

Therefore, the critical points are x = (-2 + √13) / 3 and x = (-2 - √13) / 3.

To determine the monotonicity of the function, we can use the first derivative test. We can evaluate the sign of y' in the intervals (-∞, (-2 - √13) / 3), ((-2 - √13) / 3, (-2 + √13) / 3), and ((-2 + √13) / 3, +∞).

By substituting test points into y', we can determine the sign of y' in each interval.

Let's evaluate y' at x = -3, x = 0, and x = 1:

For x = -3: y' = (3(-3)^2 + 4(-3) - 3) / ((-3) + 2)^2 = 22 / 1 = 22 Since y' is positive at x = -3, the function is increasing in the interval (-∞, (-2 - √13) / 3).

For x = 0: y' = (3(0)^2 + 4(0) - 3) / (0 + 2)^2 = -3 / 4 Since y' is negative at x = 0, the function is decreasing in the interval ((-2 - √13) / 3, (-2 + √13) / 3).

For x = 1: y' = (3(1)^2 + 4(1) - 3) / (1 + 2)^2 = 2 / 9 Since y' is positive at x = 1, the function is increasing in the interval ((-2 + √13) / 3, +∞).

Therefore, the function y = (3 - x^2) / (x + 2) is increasing in the intervals (-∞, (-2 - √13) / 3) and ((-2 + √13) / 3, +∞), and decreasing in the interval ((-2 - √13) / 3, (-2 + √13) / 3).

Extremum Analysis:

To determine the extrema of the function, we need to examine the sign of the second derivative, y''.

Let's evaluate y'' at the critical points we found earlier, x = (-2 + √13) / 3 and x = (-2 - √13) / 3:

For x = (-2 + √13) / 3: y'' = (22((-2 + √13) / 3)^2 + 36((-2 + √13) / 3) + 2) / (((-2 + √13) / 3) + 2)^4 ≈ 0.22 Since y'' is positive at x = (-2 + √13) / 3, the function has a local minimum at this point.

For x = (-2 - √13) / 3: y'' = (22((-2 - √13) / 3)^2 + 36((-2 - √13) / 3) + 2) / (((-2 - √13) / 3) + 2)^4 ≈ 0.22 Since y'' is positive at x = (-2 - √13) / 3, the function has a local minimum at this point as well.

Therefore, the function y = (3 - x^2) / (x + 2) has local minima at x = (-2 + √13) / 3 and x = (-2 - √13) / 3.

In summary, the function y = (3 - x^2) / (x + 2) is increasing in the intervals (-∞, (-2 - √13) / 3) and ((-2 + √13) / 3, +∞), decreasing in the interval ((-2 - √13) / 3, (-2 + √13) / 3), and has local minima at x = (-2 + √13) / 3 and x = (-2 - √13) / 3.