Question

1)y=x^2-x+1/x^2+x+1 2)y=ln√2x-1

Answer 1

Дивись вроді добре))

Answer 2

To analyze the given functions, let's take a closer look at each of them:

1. y = (x^2 - x + 1) / (x^2 + x + 1)

This is a rational function, which means it is a ratio of two polynomial functions. Let's analyze it further:

a) Domain: The function is defined for all real numbers except where the denominator becomes zero. To find those points, we set the denominator equal to zero and solve for x:

x^2 + x + 1 = 0

Using the quadratic formula, we get:

x = (-1 ± √3i) / 2

So, the function is defined for all real numbers except x = (-1 + √3i) / 2 and x = (-1 - √3i) / 2.

b) Asymptotes: Rational functions can have vertical asymptotes where the denominator becomes zero. However, in this case, the degree of the numerator and the denominator is the same (both are 2), so there are no vertical asymptotes.

c) Behavior as x approaches infinity: As x becomes very large in magnitude (positive or negative), the higher-order terms dominate the function, and it behaves like:

y ≈ (x^2) / (x^2) = 1

So, the function approaches y = 1 as x approaches infinity.

d) x and y-intercepts: To find the x-intercept, we set y = 0 and solve for x:

x^2 - x + 1 = 0

Using the quadratic formula, we get complex roots for x, which means there are no real x-intercepts.

To find the y-intercept, we set x = 0:

y = (0^2 - 0 + 1) / (0^2 + 0 + 1) = 1

So, the y-intercept is at (0, 1).

2. y = ln(√(2x) - 1)

This is a logarithmic function with a square root inside the logarithm. Let's analyze it:

a) Domain: The function is defined for x > 1/2, as the square root term (√(2x) - 1) must be greater than zero for the natural logarithm to be defined.

b) x-intercept: To find the x-intercept, we set y = 0 and solve for x:

0 = ln(√(2x) - 1)

There is no real solution for this equation, which means there are no x-intercepts.

c) y-intercept: To find the y-intercept, we set x = 0:

y = ln(√(2(0)) - 1) = ln(-1)

Since the square root term becomes negative, the natural logarithm is not defined for negative values, and there is no y-intercept.

d) Asymptotes: Logarithmic functions have vertical asymptotes when the argument of the logarithm becomes zero or negative. In this case, the argument (√(2x) - 1) must be zero:

√(2x) - 1 = 0

√(2x) = 1

2x = 1

x = 1/2

So, there is a vertical asymptote at x = 1/2.

e) Behavior as x approaches infinity: As x becomes very large, the dominant term inside the logarithm is √(2x). Since the square root grows slower than x, the function approaches negative infinity as x approaches infinity:

lim(x → ∞) ln(√(2x) - 1) = -∞

Please note that the graph of this function is quite complicated, so using a graphing tool might be helpful to visualize its behavior accurately.