Ask a math problem with f x x cube log2 x x squared 1 .

f(x)=x 3+log2(x+ (x 2+1)), and the domain of f(x) is r

f(-x)=-x＾3+log2(－x+√(x＾2+1))＝x＾3＋log2[1/(x+√(x＾2+1))]

x 3-log2(x+ (x 2+1))=-f(x)f(x) is an odd function.

f(x) is an increment function on (0,+).

f(x) is an increment function on r.

a+b>=0 gives a -b

f(a)≥f(-b)=-f(b)

f(a)+f(b) 0 is true.

If f(a)+f(b) 0, then f(a) -f(b)=f(-b) is known by the function being an increasing function.

A -ba + b 0 is established.

a+b>=0 is a sufficient and necessary condition for f(a)+f(b)>=0.

Finding the derivative of the function, we get that the derivative is everover to zero, so a+b and that + are the same sign, so when the denominator is zero, discuss it separately, the definition of the derivative Agree with 0|Comment: f(x) = x 3 + log2(x + (x 2+1)), where f(x) is defined in the domain r

f(-x)=-x＾3+log2(－x+√(x＾2+1))＝x＾3＋log2[1/(x+√(x＾2+1))]

x 3-log2(x+ (x 2+1))=-f(x)f(x) is an odd function.

f(x) is an increment function on (0,+).

f(x) is an increment function on r.

a+b>=0 gives a -b

f(a)≥f(-b)=-f(b)

f(a)+f(b) 0 is true.

If f(a)+f(b) 0, then f(a) -f(b)=f(-b) is known by the function being an increasing function.

A -ba + b 0 is established.

a+b>=0 is a sufficient and necessary condition for f(a)+f(b)>=0.

f(2x)=log(8xsquared + 7), then f(1)=order to clear x=1 2

The original formula f(1)=log(8 1 4+7)=log9,7, so that 2x t has a positive change x t 2

There is f(t)=log(8(t 2) 2+7> f(x)=log(2x 2+7).

f(1)=log(2+7)=log9,2,The bottom of the logarithmic function is unknown, and there is no way to solve it,2,

The title doesn't seem to be very good.

Solving the natural domain of the composite function actually requires the defined domain of f, but at the same time satisfying the range of f in order for g to make sense.

The domain of the problem g is all real numbers r, so it is only necessary to ask f to define the domain. So the definition domain of g only requires that the logarithmic function be meaningful and noisy, and the answer is all positive real numbers.

To solve the derivative of g, the function values of the points between 1" =x " =4 and the zero derivatives and the points of the state finger Zen boundary are listed and compared. The minimum value is 7 and the maximum value is 19

Substituting 1/1 of x into a known equation is.

f(1/1 of x) = (1 + 1/2 of x) divided by (1-1 of x), then the numerator and denominator are divided at the same time, the numerator becomes (x squared + 1) divided by x squared, and the denominator is (x squared - 1) divided by x squared, and then it is divided into x, and the square of x is reduced, leaving f(1/x) = (x squared + 1) divided by (x squared - 1), f(1/x) = negative (x squared + 1) divided by (1-x squared) = negative f(x).

Substituting 1 x into the equation yields f(1 x) = (1+2 x+1 x 2) (1-2 x+1 x 2).

Then multiply the top and bottom by the square of x.

f(1 x) = 1 + x squared) divided by (1-x squared) so f(1/x) = -f(x).

You only need to bring 1 x into f(x) to prove the answer:

f(1/x)=(1+1/x2)/(1-1/x2)=(x2+1)/(x2-1)=-1+x2)/(1-x2)=-f(x)

x2 represents the square of x (I know that there should actually be an input method, which can be a simple symbol letter, hehe).

All are reduced to multiples: f(x)=3logx, g(x)=4logx, and then you can see by drawing:

When 0g(x).

When x=1, f(x)=g(x).

When x>1, f(x).