Alternative method

y

=

2x

+ 1

-

(x+3 under the root number) solution: x+3=t under the root number.

Then x = t 2-3 and t >；; =0

y=2x

+ 1

-

X+3) = 2 (T2-3)+1-t = 2t2-t-5 = 2 (t-1/2) 2-5-1/2.

=2(t- 1/2)^2- 1 1/2

Because t & gt=0

Quadratic function evaluation domain

Obviously y & gt=- 1 1/2.

Therefore, the value range is [- 1 1/2, positive infinity) 2. The matching method y = x 4+2x 2-1solution: y = (x 2+1) 2-2, and the scope of the topic x is not given. If x \,

Solution: f(x)=4(x+ 1)-5.

/x+ 1

=4

-

(5/

x+ 1)

When x+ 1 >; 0, that is, x>- 1, and the range is: f (x)

When x+ 1

4. Direct method (observation method) is used for simple analytical expressions.

Y = 1-√ x ≤ 1 Solution: Range (-∞,

1]

Y = (1+x)/(1-x) = 2/(1-x)-1solution: Range (-∞,-1) \

Inequality method

6.

Maximum method

7.

Inverse function method, are these three methods necessary? It feels very useful! Hope to adopt! ! ! !