Well, you can forget about f(2), so what is f(t)? You will definitely disdain to say, just bring in x=t, and finally f(t)= 1/( 1-t), ok?
Then come on, what if it's f(2m+ 1)? Then you should know that f (2m+1) =1(1-(2m+1)) =-1/2m, right?
At this point, you know, whatever it is, I just need to replace the X in the expression after f(x) with the contents in brackets of f(x). Isn't f(f(x)) f( 1/( 1-x))? That is to say, replace the x after the equal sign with 1/( 1-x), then f (f (x)) = f (1(1-x)) =1(1).
The second problem requires calculus. I don't know if you have learned it. Look at your first question, you should still be a high school student, right?
More questions and answers?
Ask close questions
Hello, teacher. I am poor at math, but as long as you tell me, I'm sure I can understand. Teacher, please tell me the second question again. I graduated from high school for almost three years.
One answer after another.
First, you have to draw a picture. The black part is the required area. Let the required coordinates be P(t, t? )
So finding the minimum is the sum of two integrals:
f(t)=∫[0→t](t？ -y)dx+∫[t→ 1](y-t？ )dx
=∫[0→t](t？ -x？ )dx+∫[t→ 1](x？ -t？ )dx
=(t？ x- 1/3 x？ )|[0→t]+( 1/3 x？ -t？ x)|[t→ 1]
=t？ - 1/3 t？ -0+ 1/3-t？ -( 1/3 t？ -t？ )
=4/3 t？ -t？ + 1/3
Derive t so that f`(t)=4t? -2t=0, t 1=0 (excluding), t2= 1/2.
So when t= 1/2, f(t) has the maximum value, which is 4/3*( 1/2)? -( 1/2)? + 1/3= 1/4
At this time, the coordinates of point P are (1/2, 1/4).