Question

1) f(x)=x2+2x/x-12) f(x)=2cosx-tgx

Answer 1

f'(x) =[ (2x+2)*(x-1) -(x^2+2x)*1 ] / (x-1) ^2 = [ 2x^2-2-x^2-2x ] /(x-1)^2 =

=[ x^2-2x-2 ] / (x-1)^2

f'(x)= 2*(-sinx) - 1/cos^2x = -2sinx - 1/cos^2x

Answer 2

Let's analyze both functions:

1. f(x) = (x^2 + 2x) / (x - 12)

This is a rational function, which means it is a ratio of two polynomials. The function is defined for all values of x except when the denominator (x - 12) is equal to zero, as division by zero is undefined. So, to find the points where the function is not defined (the vertical asymptote), set the denominator equal to zero and solve for x:

x - 12 = 0 x = 12

So, the function is not defined at x = 12, and there is a vertical asymptote at x = 12.

To find the horizontal asymptote, we need to look at the degrees of the numerator and denominator polynomials. The degree of the numerator is 2 (highest power of x) and the degree of the denominator is 1. In this case, the horizontal asymptote is given by the ratio of the coefficients of the highest power of x in the numerator and denominator. The coefficient of x^2 in the numerator is 1, and the coefficient of x in the denominator is 1. So, the horizontal asymptote is y = 1x^2 / 1x = x.

To find any x-intercepts or y-intercepts, set f(x) equal to zero to find the x-intercepts and set x equal to zero to find the y-intercept:

x^2 + 2x = 0 x(x + 2) = 0

x = 0 or x = -2

So, the x-intercepts are at x = 0 and x = -2.

To find the y-intercept:

Setting x = 0 in the original function: f(0) = (0^2 + 2(0)) / (0 - 12) f(0) = 0 / (-12) f(0) = 0

So, the y-intercept is at y = 0.

2. f(x) = 2cos(x) - tg(x)

It seems there might be a typo in the second function. I assume tg(x) should be tan(x) (tangent function). In that case, the function becomes:

f(x) = 2cos(x) - tan(x)

This function is a combination of the cosine function (cos(x)) and the tangent function (tan(x)). Both cosine and tangent are periodic functions, and their range is (-∞, ∞).

There are no restrictions on the domain of this function; it is defined for all real values of x.

Since both cosine and tangent functions are continuous and have no vertical asymptotes, there are no vertical asymptotes for this function.

As for horizontal asymptotes, because neither cosine nor tangent has a horizontal asymptote, the function f(x) = 2cos(x) - tan(x) also does not have a horizontal asymptote.

To find x-intercepts, set f(x) equal to zero and solve for x:

2cos(x) - tan(x) = 0

Unfortunately, this equation cannot be easily solved analytically for all x values, and it would require numerical methods or graphical methods to find the x-intercepts.

As for the y-intercept, set x = 0:

f(0) = 2cos(0) - tan(0) f(0) = 2 - 0 f(0) = 2

So, the y-intercept is at y = 2.

Please note that if the expression "tg(x)" was meant to be something else, please provide the correct function, and I'll be happy to help you further.