The question of the maximum value of the function, how to find the maximum value of the function

It is known that the derivative of f(x)=(-1 3)x +bx +cx+bc is f(x), g(x)=|f´(x)|, the maximum value of g(x) on x [-1,1] is m, and if m k is constant for b and c r, find the maximum value of k.

This problem is in fact to find the minimum value of m, k

f (x)=-x +2bx+c=(b +c)-(x-b) is a parabola with a vertex u(b,b +c) with an opening pointing downward.

g(x)=|f´(x)|=|(b²+c)-(x-b)²|g(-1)=|c-2b-1|，g(1)=|c+2b-1|.

When (b + c) 0, g(x) = (x-b) -b +c), is a parabola with the vertex u(b, -(b +c)), the opening is upward, m=max, when g(-1)=g(1), that is, b=0, m obtains the minimum value, at which time m g(-1)=g(1)=1-c

When (b +c) 0, g(x) is a piecewise function:

g(x)=(x-b) -b +c), x b- (b +c) or x b+ (b +c).

g(x)=(b²+c)-(x-b)²，b-√(b²+c)≤x≤b+√(b²+c)

When g(-1)=g(1), that is, b=0, solve g(x)=(b +c), x=(2)c, if 2c 1, m=b +c=c;

If 2c 1, m=max, when g(-1)=g(1), m g(-1)=g(1)=1-c

To sum up, k=min

Solution: f(x)=(-1 3)x +bx +cx+bc then: f'(x)=-x +2bx+c=-(x-b) +b +c and g(x)=| f’(x)|=|b²+c-(x-b)²|According to the meaning of the title, no matter what the values of b and c are, if and only if b = (-1+1) 2 = 0, m has a minimum value k = |c-1|

I don't know if you can draw g(x)=| f’(x)|=|b²+c-(x-b)²|Images?

There are generally three ways to find the maximum value of a function.

The first is to take the fundamental inequality and reduce it into the form of the fundamental inequality, so that its maximum and minimum value can be found.

The second is the quadratic function, the maximum and minimum value of the six-quadratic function is characterized by the sub-formula of the required quadratic function, if the opening is up, then there is a minimum value and a maximum value of the opening downward.

The third is to take advantage of the monotonicity of functions.

The maximum and minimum value of the function can be obtained by using the definition of monotonicity or by deriving it directly.

Consider x0 to be the extreme minimum.

Then there is x0, some field o in o, any point x except x0 has f(x)>=f(x0)And there must be x1 > x0 in this critical domain, (i.e., x1 is in the right half of the critical domain) so that this inequality sign is strictly true. i.e. f(x1) > f(x0).

Otherwise, if the values of the functions in the right half of the field are all equal, and they are all f(x0), then each point is an extreme point, and thus contradictory.

Now, for x>x0, there must be f(x)>=f(x0);

Otherwise, set x2>x0, and f(x2)x0.

Now you look at the interval [, since x1 is in this interval, and f(x1)>f(x0)=f(x3), then the maximum value of f in the closed interval [x0,x3] must not be obtained at the boundary, but internally, and the maximum value on this closed interval is another extreme point other than x0, thus contradictory. (If x1 > x2 before, then x2 is in this closed range, then the minimum value is found).

Therefore, for all x>x0, there is f(x)>=f(x0);

Similar to everything x=f(x0).

Thus x0 is the minimum point.

Consider setting the interval to (a, b) and x0 to divide the interval into (a, x0), x0, b) and left and right intervals, both of which are monotonic intervals according to the title, so the maximum and minimum values of each interval are obtained at the endpoint.

If x0 is the minimum point, then.

a,x0) is a monotonic decrease, f(x0) is the minimum value in this interval (x0, b) is a monotonic increase, and f(x0) is the minimum value in this interval, so f(x0) is the minimum value of the whole interval.

If x0 is the maximum point, then.

a,x0) is a monotonic increase, f(x0) is the maximum value in this interval (x0, b) is a monotonic decrease, and f(x0) is the maximum value in this interval, so f(x0) is the maximum value of the whole interval.

Therefore, the conclusion holds.

Let the function f(x) be continuous on i (i is a finite or infinite interval) and there is only one extreme point x0 in i

Consider setting it to a minimum. Assuming that the minimum value is not obtained at the point x0, then f(x) must obtain the minimum value at the point x1, and f(x1) must be the minimum value of the function, which contradicts f(x) which has only one extreme point x0 in i.

Therefore, the assumption is wrong, i.e., x0 must be the maximum point of f(x).

The function f(x) satisfies f in the interval [a,b].'x 0, then f(x) is continuous on [a,b] and increases monotonically.

So, minf(x)=f(a) maxf(x)=f(b) for the function f(x)=sinx+cosx = 2(sinxcos 4+cosxsin 4) = 2sin(x+ 4).

When - 2 x 2 - 4 x + 4 3 4|f(x)|=√2|sin(xπ/4)|≤2-√2≤f(x)≤√2

So, the maximum value of f(x)=sinx+cos in [- 2, 2] is 2 and the minimum value is - 2

1.Functions are increment functions, so.

Maximum = f(b).

Minimum = f(a).

Maximum = f(4) = 2

Minimum value = f(-2) = -1

A function y=ax2+bx+c corresponds to a parabola, and its maximum value is divided into the following cases: First, x has no limit and can take the entire defined domain. In this case, the value of the parabola's vertex y is the maximum value of the function over the entire defined domain, that is, when x is taken as the value of the symmetry axis of the parabola, i.e., when x=-b 2a, the resulting y value is the maximum value of the function.

When a is a positive number, the parabolic opening is upward, and the resulting maximum value is the lowest point of the parabola, which is the minimum value, and there is no maximum value for this function. When a is negative, the parabolic opening is downward, the maximum value of all is the maximum, and there is no minimum value for this function. Second, given a range of variation, x can only take part of the parabola, so it is necessary to determine whether the range that x can take includes the axis of symmetry of the parabola x=-b 2a.

If it is included, then one of its maximums must be obtained at the axis of symmetry (the maximum or the minimum value is determined by the positive or negative value of a, a positive is the minimum value, and a negative is the maximum value). The other maximum value appears at the endpoint of the given defined domain, at this time, you can bring the values of both endpoints into the function, calculate the y value separately, and compare it; If the algebraic form is given, it can also be judged by the distance from the axis of symmetry, and the endpoint with the greater distance from the axis of symmetry can be taken to the maximum. If the value range of x does not include the axis of symmetry, no matter how many segments the defined domain is divided into, its maximum value must appear at the endpoint of the defined domain, and when a 0, the endpoint farthest from the axis of symmetry obtains the maximum value, and the nearest endpoint obtains the minimum value.

When a is 0, the farthest end obtains the minimum value and the farthest end obtains the maximum value. That's basically it.

It's not clear. The last thing you can do is to bring options into the question.

f(x)=x 3-ax 2-4x+4a, then f'(x)=3x 2-2ax-4, and f (-1)=0, so 3+2a-4=0, a=1 2

So f'(x)=3x 2-x-4=(3x-4)(x+1)=0, then x=4 3, x=-1

x (negative infinity, -2) -2 (-2,-1) -1 (-1,4 3) 4 3 (4 3,2) 2 (2, positive infinity).

f' >0 >0 <0 >0 >0

f(x) Increment 0 Increment 9 2 Subtraction -50 27 Increment 0 Increment.

Therefore, the maximum value of f(x) on [-2,2] is 9 2, and the minimum value is -50 27.

If the derivative is equal to 0 after the derivative of the 2nd function, it is the maximum value (a<0) or the minimum value (a>0).

If it is a higher-order function, then find the derivative of the function, and then make its derivative equal to 0 to find n points, and then divide n+1 regions according to n points, respectively, substitute any value in the region, from left to right, if the 2 adjacent regions are one positive and one negative, then it is the maximum or maximum value (the derivative is greater than 0 is the increasing function, less than 0 is the subtraction function, so it is the maximum value or maximum), if the adjacent region is one negative and one positive, it is the minimum or minimum value. If it is positive or negative, it is rounded. Finally, we bring in the two defining fields of the function x.

If ax 2+bx+c is derivatived: 2ax+b equals 0 and x=-b 2a a>0 is the minimum value and a<0 is the maximum.

The interpretation of the gods is shown in the diagram of the blind spine grip seepage in the following paragraph.

Answer: Liang Liang for the record, look at the next to destroy the picture friend socks.