Finding the answer to a function question is a bit hard to do when you first learn it

f(-x)+f(x)=0 so it's an odd function.

2) Derivative of the function.

f'（x）=[a(a^x·lna+a^(-x)·lna)（a^2-1）]/（a^2-1）^2

When 00, a (-x) > 0 so f'(x) >0, the function increases monotonically.

When a>1.

LNA>0,A2-1>0,AX>0,A(-X)>0SoF'(x) >0, the function increases monotonically.

So the function f(x) is a monotonically increasing function.

3) Informed by (2).

f(x)min=f(-1)=a(a2-1)*(a-1-a)=-1, so b<=-1 i.e. b( -infinity, -1].

1. To judge the parity of the function is to see f(-x)=f(-x) is an even function, f(-x)=-f(x) is an odd function, and -x is brought into the above equation, which can be seen as -f(x) so it is an odd function.

2.To judge the monotonicity of the function, if it is a basic function, similar to a simple function such as one, two, etc., let x1-x2>0 bring in the formula to calculate the sign of f(x1)-f(x2), if it is greater than 0, it is an increasing function, if it is less than 0, it is a decreasing function; For the above titled Transcendental Equation, it is advisable to find a derivative to determine if f'(x)>0 is the increasing function, and vice versa is the decreasing function, f'（x）=[a(a^x·lna+a^(-x)·lna)（a^2-1）]/（a^2-1）^2

When 00, a (-x) > 0 so f'(x) >0, the function increases monotonically.

When a>1.

lna>0，a^2-1>0,a^x>0,a^(-x)>0

So f'(x) >0, the function increases monotonically.

Therefore, the function f(x) is a monotonically increasing function (in this case, since the power function involves letters, the range of values of a should be discussed in the range of a).

3. Since the conclusion obtained in the second sub-problem is an increasing function, and [-1,1] is greater than b, therefore, bring in x=-1 to get f(-1)=-1, so the value range of b is smaller than -1, so it is (-1].

If you want to draw a picture, draw it yourself.

If it's a recipe. y=2x^+4-7

Isn't what you're writing right? 4 There should be x on the back, right?

y=2x^+4x-7

2（x+1）²-9

Function image: the opening direction is upward, and the axis of symmetry is the straight line x=-1 vertex coordinates: (-1, -9).

When x -1, y decreases with the increase of x, when x = -1, the function y has a minimum value of -9, and when x -1, y increases with the increase of x.

If you say yes.

y=2x²-3

The axis of symmetry should be the y-axis, x 0, y decreases with the increase of x, x 0, y increases with the increase of x, when x = 0, the function y has a minimum value of -9

1.If we know that the domain of the function y=f(x) is r, and the range is [-2,2], then the domain of the function y=f(x+1) is also r, then the range is [-2,2]2It is known that the domain of the function f(x) is defined as [a,b].

Let a x b and a -x b

then a x b and -b x -a

Because -b a 0

So a x -a

That is, the domain of the function f(x) f(x) + f(-x) is [a,-a]3If the domain of the function y=f(x) is [0,2] let 0 x+a 2 and 0 x-a 2

then -a x 2-a and a x 2+a

Because 0 a 1 is a -a, 2-a 2+a is a x 2-a

That is, the function y=f(x+a)+f(x-a) (where 0 doesn't understand, you can continue to ask, I hope it can help you.)

The value range is -2,2

The range is unchanged.

Because: f(x+1) is only a horizontal translation compared to f(x), and the specific problem is to translate f(x) one unit to the left. There is no change in the up and down direction.

So, the range is also [-2,2].

2)-b, so the -bf(x) definition domain is [a,b].

f(-x) defines the domain as -x [a,b].

Then -b then the f(x) domain is [a,b] [b,-a]=[a,-a]3)f(x+a)+f(x-a).

Everything inside must be satisfied [0,2].

0<=x+a<=2

0<=x-a<=2

a<=x<=2-a

a<=x<=2+a

Because 0 takes the intersection to get.

a<=x<=2-a

So the definition domain is [a,2-a].

1.Since the domain of f(x) is r, the domain of f(x+1) is also r, and the image f(x) is shifted 1 unit to the left on the x-axis, so the range is still [-2,2].

The definition domain should be A

If this question is changed to, knowing that the definition domain of f(x) is [0,2], how would you find the definition domain of f(y 2)?

First of all, X and X 2 are relatively independent and have no direct relationship. In the problem of functions, it is necessary to fully understand the definitions in order to do such a good job.

f(y 2) is actually a composite function, and the function g(y)=y 2 is used as an independent variable for f(x). That is, it is actually f(g(y)) that finds the domain defined when the independent variable is y. Then it is enough to have the range of g(y) in [0,2].

In a word, the value range of the inner layer of the composite function should be equal to the definition domain of the outer layer!

If you still have questions, hi me

Seeing this, the solutions to the above questions for the coming year will be clear. f(-x) actually replaces the x in f(x) with the function g(x)=-x, so just replace all the x's in the f(x) expression with -x; The same goes for f(1+x);

1. Determine the letter coefficient.

Example 1 of the copy value rangeKnowing the proportional function, then when m= and y decrease as x increases. Solution: According to the definition and properties of the proportional function, and m<0, i.e., and , so .

2. Compare the size of the x value or the y value Example 2It is known that the points p1(x1,y1) and p2(x2,y2) are two points on the image of the primary function y=3x+4, and y1 > y2, then the magnitude relationship between x1 and x2 is ( )a x1>x2 b.

x10, and y1>y2. According to the nature of the primary function, "when k > 0, y increases with the increase of x", x1 > x2 are obtained. So choose A.

3. Example 3 of judging the position of the function imageOnce the function y=kx+b satisfies kb>0, and y decreases with the increase of x, then the image of this function does not pass through ( )aFirst quadrant b

Second Quadrant CThird Quadrant dFourth Quadrant Solution:

From kb>0, k, b with the same sign. Because y decreases with the increase of x, k < 0. So b<0.

Therefore, the image of the primary function y=kx+b passes through the first.

The second, third, and fourth quadrants do not pass through the first quadrant. So choose A.

Let this line be y=kx+b

mk+b=n①

nk+b=m②

m-n)k+b=n-m③

Get (m-n)k=n-m

b=0k=-1

The function passes. 2. Four quadrants.

Solution: Bring f(x) into the following inequality to obtain:

a^3-ta-ln(√(a^2+1)-a)+b^3-tb-ln(√(b^2+1)-b)-ln(√(a^2+1)-a)-ln(√(b^2+1)-b)。

Take the previous minus sign in, and a+b is not equal to 0; Divide the sides by a+b

1) When a+b>0:

t>/a-(-b)

Here g(x)=ln[ (x 2+1)+x], which can be seen as a point (a,g(a)); b,g(-b)) is a straight line formed by two points.

For the slope of the x-axis, the slope of any two points on the curve can always be expressed by the tangent slope of one point on the curve at the changing point (that is, the Lagrangian median theorem to be learned in college).

So t>max g(x).'Finding the derivative of g(x) gives it a derivative of 1 (x 2+1).Note that x is not equal to 0 here, otherwise a=b=0. So t can be taken to be equal to 1.

then t>=1

2) When a+b<0, because you start to multiply a 3 + b 3 by it, it will be reversed "for", so then divide a + b and follow the above.

So here there is also t>=1.

In summary: t>=1

Take a=0,x>0, and have f(x)(ln(sqrt(1+x 2)+x) xWhen x tends to 0, using Lopida's rule to get t>=<0 is similar. So t = 1.

(1+tanhx) (1-tanhx)=e (2x) log[(1+tanhx) (1-tanhx)]=2x, so the inverse function of tanhx=y is x= (1 2) log [ 1+y) (1-y)].

2) tanhx= sinhx coshx, so tanh(x+y).

sinh(x+y)/cosh(x+y)

e^(x+y)-e^(-x-y)]/[e^(x+y)+e^(-x-y)]

tanhx+tanhy)/(1+tanhxtanhy)=/=/

2[e (x+y)-e (-x-y)] = [e (x+y)-e (-x-y)] [e (x+y)+e (-x-y)], so the equation holds.