Math Help Plz.. : )?

Question

I can't seem to understand this and my calc is computing right.

a.)The max and min values of f(x)=l9-x^2l where -2
b.) Find the Max and min values of f(x)=xsqr(2-x)

c.)Find the x coordinates of points of inflection of f(x)=x+sin x.

Help on any of these problems would me most grateful.

Answer 1

A) Because -2 So -7<9-x^2<9, So 0<|9-x^2|<9, So 0
B) xsqr(2-x)=sqr{x^2(2-x)}, So x^2(2-x)>=0, So 2-x>=0
So x<=2; x^2(2-x)=2x^2-x^3, So 2x^2-x^3>=0, So f(x)>=0
there isn't max value

C) x+sinx=0, sinx=-x, Because -1<=sinx<=1, So -1<=x<=1
So x=0

Answer 2

A) a table of values will help you see the pattern:
x | f(x)
-2|5
-1|8
0|9
1|8
2|5
3|0
4|7
Thankfully, the highest and lowest points are integers, so they're shown in this table. highest=9, lowest=0
B)This is confusingly written... is it X times the root of (2-X)?
C)Pt of inflection is when 2nd derivative is 0
f(x)=x+sinx
f'(x)=1 + cos x <--derive of x is one, derive of sin is cos
f''(x)=0+ (-sinx)=-sinx
0=-sinx <== let the 2nd derive equal 0. when solved, x= pt of inflection.
this can be multiplied by -1
0=sin x So whenever sin x is zero, then there'll be a pt of inflection: x=0, PI, 2Pi, 3PI (0 degrees, 180 deg, 360deg...)

Answer 3

a) for -3
f(-2)=-9+(-2)^2=-5 min

f(4)=9-16=-7 max


b) find y' then y'=0 from here x=?
y '=1*sqrt(2-x)+x (-1/2sqrt(2-x))

c)

y '= 1+cosx

y ''=-sinx

if y''=0 then (x,y) is inflection point

-sinx=0

or

sinx=0
x=0 or pi