Function questions won t yes function questions.

f(x)=a(x-1 a) 2-1 a+a+b1) when a > 0, if 1 a belongs to [0,3], the minimum value is f(x)=-1 a+a+b=1 when x=1 a

The maximum value is taken when x = 0 or 3, i.e. a + b = 5 or 9a - 6 + a + b = 5 with a system of equations: -1 a + a + b = 1

a+b=5 a=1 4 , b=19 4 rounded off because a>1 3 or: -1 a+a+b=1

9a-6+a+b=5 a=1,b=1 a=1 9 rounded.

2) When a > 0, if 1 a does not belong to [0,3], that is, a belongs to [0,1 3], then 1 a > 3, and the point where the extreme value is taken is 0,3, and it can be seen from the image that on the left side of the axis of symmetry, the function is decreasing, so f(0)=5 f(3)=1

a+b=5, 9a-6+a+b=1 a=2 9,b=43 93) when a<0, 1 a <0

From the function image, we can see that when x=0 takes the maximum value of 5, when x=3 takes the minimum value of 1, i.e. a+b=5, 9a-6+a+b=1 a=2 9, b=43 9 rounded.

So a=1, b=1 or a=2 9, b=43 9

Classification discussion: The axis of symmetry of a quadratic function image is x=1 a1, when 0<1 a<=3, i.e., a>=1 3

Reclassification, (1) when x=0 takes the maximum value and solves a=1 4 (rounded) 2) when x=3 takes the maximum value and solves a=1, b=12, when 1 a>3, that is, a<1 3

x=3 is the minimum.

x=0 is the maximum.

The solution yields a=2 9 b=5-2 9

3. When 1 a<0 is a<0

x=3 takes the maximum value.

x=0, take the minimum value a=10 9 and round it off.

Sum up.

ice110 got it right. Don't look closely. One look at the discussion of the axis of symmetry and you know that it is right. Something else is definitely wrong.

(1+tanhx) (1-tanhx)=e (2x) log[(1+tanhx) (1-tanhx)]=2x, so the inverse function of tanhx=y is x= (1 2) log [ 1+y) (1-y)].

2) tanhx= sinhx coshx, so tanh(x+y).

sinh(x+y)/cosh(x+y)

e^(x+y)-e^(-x-y)]/[e^(x+y)+e^(-x-y)]

tanhx+tanhy)/(1+tanhxtanhy)=/=/

2[e (x+y)-e (-x-y)] = [e (x+y)-e (-x-y)] [e (x+y)+e (-x-y)], so the equation holds.

If you want to draw a picture, draw it yourself.

If it's a recipe. y=2x^+4-7

Isn't what you're writing right? 4 There should be x on the back, right?

y=2x^+4x-7

2（x+1）²-9

Function image: the opening direction is upward, and the axis of symmetry is the straight line x=-1 vertex coordinates: (-1, -9).

When x -1, y decreases with the increase of x, when x = -1, the function y has a minimum value of -9, and when x -1, y increases with the increase of x.

If you say yes.

y=2x²-3

The axis of symmetry should be the y-axis, x 0, y decreases with the increase of x, x 0, y increases with the increase of x, when x = 0, the function y has a minimum value of -9

f(1)+f(1)=f(1) gives f(1)=0f(1 x)+f(x)=f(1)=0, so that f(1 x)= f(x) sets x2>x1>0 , x2 x1 > 1, f(x2 x1)>0

f(x2)-f(x1)=f(x2)+f(1/x1)= f(x2 / x1)>0

f(x2) >f(x1)

f(x) is an increasing function on (0, positive infinity).

Proof: According to the title, x2>x1>0 can be set so that x2 x1=k>1 (k must be greater than 1).

Then f(x2)-f(x1)=f(kx1)-f(x1)=f(k)+f(x1)-f(x1)=f(k)>0 (set by the question).

So when x2>x1>0, f(x2) > f(x1), i.e., the increment function.

by f(2)=1

Let x1=2, x2=1

then f(2) = f(2) + f(1).

Therefore f(1) = 0

Again when x 1, f(x) 0

So f(x) is an increasing function over (0, positive infinity).

<> as long as the value of m and n is satisfied, the group is only around. The spring of the clan can collapse and cultivate.

Let 2 x=t (1 2,8).

So y=log2(t 2-t).

When t=8 has the maximum log2(56)=3+log2(7).

Take the function in parentheses at the highest point of (-1,3) and substitute log( ) to calculate, and find the maximum value of f(x).

1) Since the abscissa is k, the destroyed thought is the coordinate of the zero point of a (0,1).

2) The ordinates of A1, B1 and Lusen C1 are all 2, the abscissa of A1 is 0, the abscissa of B1 is K 2, and the abscissa of C1 is K;

The ordinates of A2, B2, and C2 are all 3, the abscissa of A2 is 0, the abscissa of B2 is K 3, and the abscissa of C2 is K; Slim sleepy.

a1(0,2 )b1(k 2,2) c1(k,2) then c1b1=k 2 2+4 ; a1b1=4;Then c1b1 a1b1=k 2 8+1=k 2 (2 3)+1

a2(0,3 )b2(k 3,3) c2(k,3) then c2b2=k 2 3+9 ; a2b2=9;Then c2b2 a2b2=k 2 27+1=k 2 (3 3)+1

an(0，n+1） bn(k/(n+1),n+1) cn(k,n+1)

then there is cnbn=k 2 (n+1)+(n+1) 2,anbn=(n+1) 2; Then cnbn anbn=k 2 (n+1) 3+1