What is the maximum value of the function f x x squared minus 2X plus 3 over the interval minus 2,3

Solution: f(x)=x -2x+3

x-1)²+2

The vertex is (1,2), and the axis of symmetry is x=1,2 from the axis of symmetry =1-(-2)=33 from the axis of symmetry = 3-1=2, and it can be seen from the image that the value y is the largest at x=-2, so the maximum value is =(-2-1) +2=11

This question can also be understood in this way:

The f(x) vertex is (1,2), the axis of symmetry is x=1, and the opening is upward, so the maximum value must appear on both endpoints, just compare the size of the two endpoints.

f(-2)=11；f(3)=6, so the maximum value is 11

f(x) = square of x minus 2x plus 3 = square of (x-1) + 2

The maximum value over the interval minus 2,3 is when x=-2.

f(x) = square of x minus 2x plus 3 = square of (x-1) + 2 = 11

Hello, f(x)=-2x 2+3x-1=-2(x 2-3x 2)-1=-2(x-3 4) 2-1+9 8=-2(x-3 4) 2+1 8 we can see that the parabolic vertex is (3 4,1 8) and the opening is downward, so we can know that on the right side of the vertex, that is, when x>3 4, the function is a decreasing function, and when x<3 4, the function is an increasing function, so the function is a decreasing function on the interval [2,5]. So when x=2, the value of f(x)=-2x 2+3x-1 is the largest (the rest is calculated by yourself).

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Function f(x) = x -2x+3

x-1)²+2

The combination of numbers and shapes can be known:

When -2 x 3, f(x)max=f(-2)=11

f(x)=3x^3-9x+5 f'(x)=9x 2-9>0 x1 So the monotonically increasing interval of f(x) is (-infinity, -1) and (1,+infinity) f(-2)=-1 f(-1)=11 f(1)=-1 f(2)=11, therefore, f(x) is 11 and the minimum is -1 when [-2,2] goes up the hill between the caves of the district

f(x)= x 2+2x+1 is the number of increments in the interval [-3,a], the parabola opening is downward, and the axis of symmetry is the trace thickness: x=1;

The increasing interval of the parabola is: (-1];

According to the title [-3,a] must be a subset of (-1), ie. a≤1

f(x)=x^2-4x+3

x-2) 2-1, parabolic opening magna upward, symmetry axis hand cherry tree x=2, minus interval (-infinity, 2 stuffy potato book.

f'Hail ruler (x) = 3 x 2-6x-9 = 0

x1=-1 x2=3

f(0)=2

f(4)=-18

f(-1)=7

f(3)=-25

The minimum source trapped high is -25, and the maximum is 7

Pure mouth counts the square foot.

f(x)=x 2-3x+2=(x-3 2) 2-1 4, because x is sensitive to [1 2,2], so f(x) is monotonically decreasing in the interval [1 shouting or 2,3 2];

The maximum value of f(x) is Zheng Wu, when x=1 2, then f(x)=3 4

The maximum value is at -2 because the axis of symmetry is 1 which is a concave function -2 is far away, so the maximum value at -2 is 11

f(x) = square of x minus 2x plus 3 = square of (x-1) + 2

The maximum value over the interval minus 2,3 is when x=-2.

f(x) = square of x minus 2x plus 3 = square of (x-1) + 2 = 11

Function f(x) = x -2x+3

x-1)²+2

The combination of numbers and shapes can be known:

When -2 x 3, f(x)max=f(-2)=11

f(x)=3x^3-9x+5 f'(x) = 9x 2-9>0 x<-1 or x>1

So the monotonically increasing interval of f(x) is (-infinity, -1) and (1, + infinity) f(-2)=-1 f(-1)=11 f(1)=-1 f(2)=11

Therefore, the maximum value of f(x) over the interval [-2,2] is 11 and the minimum is -1

f'(x)=9x^2-9=0

x1=-1,x2=1

The monotonic increase interval is (-infinity, -1) and (1, positive infinity) f(-2)=-1

f(-1)=11

f(1)=-1

f(2)=11

So the maximum value is 11, and the minimum value is -1

f'(x)=9x 2-9 reamf'(x) >=0 gives x>=1 or x<=-1, so the monotonically increasing interval is (-1] and [1,+

On [ on x=-1 is the maximum f(-1)=11, x=1 is the minimum f(1)=-1, and f(-2)=-1, f(2)=11So the maximum value is 11 and the minimum value is 1

A is equal to zero. The minimum value of the number sell shout is 3

A 1, the minimum value is 4-2a

A -1, the minimum value of repentance is 4 plus 2a

At 1 a 1, the minimum value is 3-a flat potato field.