High school math extremum questions, thank you in advance.

Recipe first. y=√[(x-3)^2+1]+√x+3)^+4^2]

Method 1: Know that the straight-line distance between (x1,y1) and (x2,y2) in the coordinate system is [(x1-x2) 2+(y1-y2) 2]).

Think of it as a person's journey from point A (3,1) to the x-axis (which can be any point) to point B (-3,4).

Now ask for the shortest distance for this man. (Is the reflection of light learned?) Something like this).

Make point b(-3,4) with respect to point b symmetrical on the x-axis'(-3,-4)

Use the two-point formula to find the straight line ab'analytical.

l=5/6x-3/2

You can know the ab at this time'The distance is the shortest, then just find AB'The intersection with the x-axis, we know when x has the shortest distance (the minimum value of the function).

Let l=0, and the solution is x=9 5

So when x=9 5.

The function has a minimum value.

Substitute ymin= 63 (ymin represents the minimum value of y).

Method 2: Recipe y= [(x-3) 2+1]+ x+3) +4 2])).

As mentioned above, use vectors to do, let the vector oa=(3-x,1), vector ob=(x+3,4).

Because in the end, the vector OA and the vector ob should be in the same direction, so let the ordinates of both be positive (both are negative, as long as they are in the same direction), and let the modulus length of the result of the addition of vector OA and vector ob be constant, so give a negative sign to the abscissa of vector OA and become 3-x, not x-3), then there is y=|oa|+|ob|

inequalities according to vectors.

Vector a|+|Vector b|>=|Vector a + vector b|

If and only if vector a is in the same direction as vector b), take the equal sign).

oa|Represents the modulo of the vector oa">="Indicates greater than or equal to).

Yes. y=|oa|+|ob|>=|Vector oa + vector ob|

And vector oa + vector ob = (6,5).

So |Vector oa + vector ob|=√(6^2+5^2)=√63

If and only if the vector OA is in the same direction as the vector OB and the vector OA + vector OB is constant, the minimum value is taken and the minimum value is .

Vector oa + vector ob|=√63

Therefore the minimum value of y is .

ymin=√63

Give me an email and I'll send you the answer.

Do it with vectors. The answer is 5

Assumptionsf(x) and g(x) at x=aContinuous and second-order derivable, there is.

f'(a)=g'(a)=0，f''(a)<0，g''(a)<0

f'(x)=[f(x)·g(x)]'f'(x)g(x)+f(x)g'(x)

Hence f'(a)=f'(a)g(a)+f(a)g'(a)=0

So f(x) is obtained at x=aExtremum, option c is wrong.

f''(x)=[f'(x)g(x)+f(x)g'(x)]'f''(x)g(x)+f(x)g''(x)+2f'(x)g'(x)

Hence f''(a)=f''(a)g(a)+f(a)g''(a)+2f'(a)g'(a)=f''(a)g(a)+f(a)g''(a)

Only know f''(a)<0，g''(a) < 0, while the signs of g(a) and f(a) cannot be determinedSo f''The symbol of (a) cannot be determined。So, even if we know that f(x) and g(x) are second derivative at x=a, and the second derivative.

Less than zero, tooIt is only possible to determine that f(x) achieves an extreme value, but it is not possible to determine whether it is a maximum or a minimum valueSelect D correctly

Pick D, not sure.

For example, f(x)=-x 2 and g(x)=-x 2 take the maximum value at x=0, but f(x)g(x)=x 4 take the minimum value at x=0.

For example, f(x)=2-x 2, g(x)=2-x 2 takes the maximum value at x=0, and f(x)g(x)=(2-x 2) 2 takes the maximum value at x=0.

Because the function f(x, y) has a zero first derivative (slope) at its extreme (limit) point. From this, the extreme point can be obtained by finding the function to solve the derivative and make it equal to zero;

f(x,y)=2x^3 - 6xy^2 + 2y^4∴ əf/əx = 6x^2 - 6y^2; əf/əy = -12xy + 8y^3.Ready:

x^2 - y^2 =0

3x + 2y^2 =0

There are three extrema for simultaneous solving: (0,0) and (3 2,3 2) and (3 2,-3 2).

Find the maximum and minimum distance from the point on the curve x -xy+y =1 (x 0, y 0) to the coordinate origin.

Solution: Let the coordinates of the moving point m on the curve be (x,y); its distance from the origin l= (x +y); Now ask for the moving point m in.

The maximum and minimum value of l when moving on a curve. To simplify the operation, the maximum and minimum values of l =x +y are found instead. Therefore it can be used to pull.

Grangian multiplier;

function f(x,y)=x +y + x -xy+y -1);

Let f x=2x+ (3x -y)=0....

f/∂y=2y+λ(x+3y²)=0...

x³-xy+y³-1=0...

Three-formula simultaneous solution obtains: x=1, y=1, =-1;

Therefore lmax= 2;When y=0 there is x =1, i.e. x=1;Therefore, the point (1,0) is the point on the curve that is on the x-axis and closest to the origin, where lmin=1;When x=0 there is y =1, i.e. y=1;Therefore, the point (0,1) is the point on the curve that is on the y-axis and closest to the origin, where lmin=1;Therefore lmax= 2;lmin=1.

Theorem 1 is a necessary condition, and by finding the derivative, it is calculated that the point where the derivative is zero failure, i.e., the stationary point.

The theorem Chabo2 is a sufficient condition, and the discriminant formula is calculated, and when > 0, there must be an extreme value; Then the stations obtained in the first step are substituted into the discriminant test one by one.

In this discussion of the extreme value problem, for the independent variable of the function, there is no other condition except that it is limited to the defined domain of the function, which is called unconditional extremum. Therefore, when =0 there is no way to tell if there is an extremum. If more conditions are given, you can continue to judge.

Also, note the difference between "extremum" and "maximum", "minimum", and "maximum". Silver Boom.

So the deep vv is not hungry, oh love from where to counterattack to the uncle and the Internet café who died in a hurry 7 me; With a baby.