High School Math Problem Function f x 2x root number 1 6x maximum value

The maximum value of the function f(x)=2x+ (1-6x).

Solution: Define the domain: from 1-6x 0, the domain can be defined as x 1 6

Let f (x)=2-6 [2 (1-6x)]=2-3 (1-6x)=0 give (1-6x)=3 2;The square root number is 1-6x=9 4, so the standing point x=-5 24

When x<-5 24 f (x) > 0;When x<-5 24 f (x) > 0;Therefore x=-5 24 is the maximum. Maximum = f (-5 24) = -(5 12) + 1 + 5 4).

When x=1 6, i.e., at the right end of the defined domain, f(1 6)=1 3;

When x is infinitely close to the left endpoint of the defined domain, x - limf(x)=- ;

Therefore f(-5 24) = 13 12 is the maximum value of the function in its defined domain.

Method 1: f(x)=-(1 3)*[sqr(1-6x)-3 2] 2+13 12>=13 12

Therefore, the maximum value is 13 12

Method 2: Let t=sqr(1-6x), then t 0f(x) is converted to g(t)=-(1 3)*t 2+t+1 3=-(1 3)(t-3 2) 2+13 12>=13 12

Therefore, the maximum value is 13 12

Method 3: Find the derivative (omitted).

Let t = root number 1-6x (t 0).

then t = 1-6x

x=1-t²/6

Original formula = 2 (1-t 6) + t

1-t²+3t/3 (t≥0）

The following is a general method of evaluating the range to find the maximum value.

The commutation method makes t = root number 1-6x to get x = 1/6 of the square of -t f(t) = -3/3 of t squared + t + 1/3 of the quadratic function and then the property of the quadratic function can be obtained to obtain the maximum value of 13/12

The domain of the easy-to-know function f(x) is [1,+infinity), and the barter function f(x) is an increasing function, so the minimum value of the number of empty acres f(x) is f(1)=1

y=x is an increment function barrier on the defined domain.

y=- 1-2x) is an increment function on the defined domain.

y=x- (1-2x) is an increment function on the defined domain.

Define domains 1-2x 0

x 1 Jane Rolling Front 2

When x=1 2, y has a maximum value, which is 1 Bizen 2

There is no minimum.

y=root number omission x 2-2x+2 + root number x 2-4x+8=root number((x-1) 2+1) + root number((x-2) 2+4) Geometric meaning: y represents the sum of the distance from point p(x,0) to point a(1,1) and the distance to point b(2,2) on the x-axis. Now is the minimum required for the sum of these two distances!

P is on Zheng Li rotten x-axis disturbance or, not on AB ,..

x=t≥0y=t-t^2

t-1/2)^2+1/4

When t=1 2, that is, x=1 4.

Y has a maximum value of 1 and 4

The maximum value of the function f (grip x) = root number x-x (x 0) is 1 4

x increases, -x subtracts, 2-x subtracts, 2-x subtracts under the root number, 2-x increases under the root number, and x-1-2-x increases under the root number, so it is an increase function.

Because the root number must be greater than or equal to 0, 2-x>=0, x<=2

So the maximum value is when x=2 and f(x)=1

The inverse proportional relation is that as x increases and y decreases, but the relationship between x and y cannot be accurately obtained by the formula. Functions can, for example, give you a value of x, y is deterministic.

f(x) increases with the increase of x, and 2-x must be greater than 0, then x is less than 2, then when x=2 there is a maximum value, which is 1

Simplified to root number x 2-x 4

x 2-x 4 is derived as 2x-4x 3=0 x = 2 out of 2 root number 2 The original function increases on 0-2 out of 2 root number 2 and decreases on 2-1 root number 2 out of 2, so there is a maximum value of 1 2 when x = 2 out of 2 root number 2

f(x)=|x|√(1-x^2)

1-x^2≥0

1≤x≤1f(-x)=|x|√(1-x^2)=f(x)f(0)=0|

f(x)=|x|(1-x 2) is an even function.

Because -1 x 1, let x=sina cosa 2=1-x 2f(x)=|x|√(1-x^2)=|sina|√(cosa^2)=|sina|*|cosa|=1/2|sin2a|

0≤1sin2a|≤1,0≤1/21sin2a|≤1/2f(x)=|x|The maximum value of (1-x 2) is 1 2

f(x) √3x(8 - x)]

3(x² -8x)

3(x - 4)²+48]

Because it was 0 < x < 5 before the burner

So when the stupid clear x = 4 the maximum value of the time zone is 48, i.e. 4 3 so the maximum value is 4 3

Matching method: f(x)=Genhao Sanchun Yunai-3(x-4) +48

At 0< x<5, take x=4 to have the maximum value, =4 times the root number to dig 3< p>

Answer: y=x is an increment function on the defined domain.

y=- 1-2x) is an increment function on the defined domain.

y=x- (1-2x) is an increment function on the defined domain.

Define domains 1-2x 0

When x 1 2x = 1 2, y has a maximum value, which is 1 2

There is no minimum.

f(x)=x-√(1-2x)

The domain is defined as 1-2x>=0, that is, x<=1 2

Let t= (1-2x)>=0, get: x=(1-t 2) 2f=(1-t 2) 2-t=1 2*(-t 2-2t+1)=-1 2*(t+1) 2+1

Since t>=0, the maximum value of f is fmax=1 2 when t=0 (i.e., x=1 2).

When t is infinity, the minimum value of f is obviously negative infinity.