Math problems with power functions, several math problems about power

1) It can be divided down to think that the 6 to the 11th power is equal to the 6th power and then the root number 11 to the 6th power is less than the 6th power. (Think of it this way, after the 6th power, the 6th power multiplied by 6, and the 6th power multiplied by 7).

And then at the same time the root number 11According to the function y=x (1 2), we can see a monotonic increase, and in the same way, y=x (1 11) should also be monotonically increased. Because the 6th power is larger than the 6th power, the 6 11th power should be larger than the 6 11th power.

Verified with a computer.

The 6 to the 11th power is.

The 6 to the 11th power is.

The second question can be solved in the same way.

Proof seems to be sufficient with a definition.

The exponents of the power function are equal, and the larger the base, the greater the value".

Keeping this in mind, it is convenient to do the problem, so (1) the 6 11 power is less than the 6 11 power 2) the 5 3 power is greater than the 5 3 power.

Proof : (1) Because f(x) = x 3, f(-x) = -x 3f(x) + f(-x) = x 3-x 3 = 0

So the function is odd.

2) Set x1 > x2

then y1-y2=x1 3-x2 3=(x1-x2)(x1 2+x1x2+x2 2).

Since x1 > x2, x1-x2>0, x1 2+x1x2+x2 2>0 is y1-y2>0, so the function y=x 3 is odd and incremental on r.

That's all, what's a good way to tell me.

In fact, it's 16 squared, 256

You can set it to x=13

n=1 and the answer is 12

2 pcs. Draw the graph and see how many points there are.

x^2 -2= 2^x

Let f(x)=y=x

Solution: Definition of even function: f(-x) = f(x).

f(-x)=(-x)^α=(-1)^αx^αf(x)=x^α

f(x)=f(-x), i.e., (-1) x =x, i.e., x [1) 1]=0

-1) 1=0, i.e. (-1) =1 is a positive even number or 0 Note: If =0, then define the x≠0( ) solution in the domain:

When >0, f(0)=0 =0, so f(x) must have an intersection with the x-axis and be rounded.

When =0, f(x)=x 0=1, so f(x) has no intersection with the x-axis. Note: x≠0 here

When ", f(x)=1 x (-so f(x) has no focus on the x-axis.

In summary: The value of (-00,0】

1.Even.

2.i.e. x cannot be equal to 0 and alpha cannot be a fraction i.e. alpha must be an integer.

The negative exponential power of a number is equal to the reciprocal of its positive exponential power.

Not really! The minus third of a is the reciprocal of the third of a, i.e. 1 (a 1 3).

No, it's a part of a under the third root number.

One-half power of 16 = 4;Negative power of half of 16 = power of half of 16, the reciprocal of half of it. It is equal to 1 4, which is a quarter.

You'd better not say "power" in the future, that refers to the concept of square, cube; It's better to say "power". This is the true language of mathematics. "Power" refers to the large cloth that covers everything.

It's like a bunch of headcloths on the heads of ethnic minorities. What we're talking about here is this little exponent, which "controls" the value of the entire number.

Obviously, the exponent is a fraction, and the denominator is an "odd number", which means that the square number to be opened is negative, so it can be "opened".

The minus 7 power of a is equal to the reciprocal of (the 7th power of a), which is ".One in one".

(1) Make the function f(x)=(m2-3m+3)x (m+2) a power function, and the function f(x) is an even function.

m^2-3m+3=1,(m-1)(m-2)=0,m1=1,m2=2

m+2=2k, (k is an integer), m=2(k-1).m1=1 (undesirable, rounded) Only m2=2 is a power function.

The value of m is 2

2) It is known that y=(m2+2m-2)x [1 (m2-1)] 2n-3 is a power function, m2+2m-2=1,1 (m2-1)≠1,(m2-1)≠0,2n-3=0

The solution is that m1=1, m2=-3, m3≠1, m4≠-1, m5≠ 2, m6≠- 2, n=3 2

m=1 (disagree, rounded), only m=-3, n=3 2, is the solution of the power function.

a to the third of a power under the root sign.

(1)f(x) 3 sinxcosx+cos2x= 3 2 sin2x+ 1 2 cos2x+ 1 2 =sin(2x+ 6 )+1 2 The minimum positive period of the function t= 2 2 = , 1 sin....

Enter or copy-paste this formula.

or =(,

1.First of all, m -2 m-3 is less than 1 when m -m-1 is greater than 0, m -2 m-3 is greater than 1 when m -m-1 is less than 0, and then the equation is solved.

2.Tell me for yourself.

1) is a power function.

m^2-m-1=1

m-2)(m+1)=0

m=2 or m=-1

2) is a power function and is a condition of the increasing function (1) on (0,+) and -5m-3>0

Get: m=-1

3) is a proportional function.

m2-m-1 does not = 0 and -5m-3 = 1

Get m = -4 5

4) is an inverse proportional function.

m2-m-1 does not = 0 and -5m-3 = 1

Get m=-2 5

5) is a quadratic function.

m2-m-1 does not = 0 and -5m-3 = 2

Get m = -1 5

f(x)=x^a

f (cube on both sides.)

Take the reciprocal 8 a=5 (-1 2).

So 1 f(8) = 1 8 a = 5 1 2 = 5 is approximately equal to.