`ararrp, brarrq, r, crarrs, r,drarrq,r.`
a. `f(1-x)=f(1+x)`
`therefore" "-f'(1-x)=f'(1+x)`
Hence, graph of f(x) is symmetrical about point (1, 0) [ as if f(x) = - f(-x), then f(x) is odd and its graphs is symmetrical about (0,0). Now shift the graph at (1,0)].
b. `f(2-x)+f(x)=0`
`"Replace" x by 1+x. Then f(2-(1+x))+f(1+x)=0`
`"or "f(1-x)+f(1+x)=0`
`"or "-f'(1-x)+f'(1+x)=0`
`"or "f'(1-x)=f'(1+x)" (1)"`
Therefore, graph of f'(x) is symmetrical about line x=1.
Alos, put x=2 in (1). Then f'(-1)=f'(3).
c. `f(x+2)+f(x)=0" (1)"`
Replace `x by x + 2. then f(x+4)+g(c+2)=0" (2)"`
From (1) and (2), we have f(x) = f(x+4)
Hence, f(x) is periodic with period 4.
Also, f'(x)=f'(x+4).
Hence, f'(x) is periodic with period 4.
Put `=-1 in f'(x)=f'(x+4). Then f'(-1)=f'(3)`.
d. `"Putting "x=0, y = 0, we get 2f(0)+{f(0)}^(2)=1`
`"or "f(0)=sqrt(2)-1" "[because f(0)gt0]`
Putting `y=x, 2f(x)+{f(x)}^(2)=1`
Differentiating w.r.t. x, we get
`2f'(x)+2fcdotf'(x)=0`
`"or "f'(x){1+f(x)}=0`
`"or "f'(x)=0," because "f(x)gt0`.
