Find the maximum value of the function f x root x 4 3x 2 13 root x 4 x 2 1

Solution: f(x) = root number (x 4-3x 2+13) - root number (x 4-x 2+1) let x 2 = t

Then the original formula = root number (t 2-3t + 13) - root number (t 2-t + 1).

Root number [(t-3 2) 2+43 4] - root number [(t-1 2) 2+3 4].

Root number [(t-3 2) 2+(0-root number 43 2) 2]-root number [(t-1 2) 2+(0-root number 3 2) 2].

Notes: (0-root43 2) 2=(-root43 2) 2=43 4, (0-root3 2) 2=(-root3 2) 2=3 4

where "root number [(t-3 2) 2+(0-root number 43 2) 2]" can be expressed as the distance between the point (t, 0) and the point (3 2, root number 43 2);

The root number "[(t-1 2) 2+(0-root number 3 2) 2] " can be expressed as the distance between the point (t, 0) and the point (1 2, root number 3 2).

Then the point (t, 0), the point (3 2, the root number 43 2), and the point (1 2, the root number 3 2) can be regarded as the three vertices of the triangle.

The difference between the two sides of the triangle is less than the third side, so the root number [(t-3 2) 2+(0-root number 43 2) 2]-root number [(t-1 2) 2+(0-root number 3 2) 2] is less than the distance between the point (3 2, root number 43 2) and the point (1 2, root number 3 2). That is, less than the following number [(3 2-1 2) 2 + (root number 43 2 - root number 3 2) 2] = (25 - root number 129) 2

So find (25-129) 2 is the maximum value of the original formula.

Addendum: Then t=?, can you get this maximum?

The point (3 2, root number 43 2) is connected to the point (1 2, root number 3 2).

The expression for the straight line is: y=[(root number 43 - root number 3) 2]x - (root number 43-3 times root number 3) 4

Its intersection with the abscissa axis is (17-129) 40,0, which is the point where the original formula obtains the maximum value of (t,0).

Because the three vertices are in a straight line, the difference between the two sides is equal to the third side

Emphasis" Trace these three points, triangles, and answers on the coordinate axis, and it will be clear at a glance.

Let t=x f(x)=g(t)=root(t+13)-root(t-t+1).

Root number [(t-3 2) +43 4] - root number [(t-1 2) +3 4] is the maximum value of the distance difference between point (t, 0) and point a (3 2, root number 43 2), and point b (1 2, root number 3 2).

The difference between the two sides of the triangle is less than the third side, so when taking the maximum value, the above three points are collinear, the maximum value is ab distance, find it yourself, and then verify t>=0...

What, the result is more complicated, are you sure you didn't make a mistake?

Root number x 2+1 plus root number (4-x) 2+4 can be seen as.

The sum of the distances from a(x,0) to b(0,1) and c(4,2) is hungry.

then f(x)=|ab|+|ac|The minimum value of rotten is positive.

Take A'（0,-1）

So f(x) is a tease =|ab|+|ac|》=a'c|=5

Then t 4 = 2

The tick function x+1 x is incremented when x>1.

So here t=2

The minimum value is 2+1 2=5 2

Continue to rewrite equivalently to:

f(x)=√x-1）^2]+（0-0）^2+√[x+4）^2+（0-3）^2]

Now let's see if the first root number is equivalent to the distance between the point (x,0) and the point (1,0) in analytic geometry; (This is well understood);

The second root is not the same as the distance between the points (x,0) and (-4,3).

The idea of the answer came out:

A moving point c(x,0), i.e. a moving point on the x-axis, is the sum of the distances between the fixed point: a(-4,3), b(1,0).

Now we find the minimum value of f(x)= x-1) 2+ (x+4) 2+9, that is, the minimum value of the distance between the moving point and the two fixed points.

Q: When is it smallest? Please see that the moving points are at points c and c'. ab

Analysis: Because x≠0, then Qisheng 1+x 2+x 4>1+x 4>0 so Quiet Bridge (1+x 2+x 4) > 1+x 4) >0 i.e. (1+x 2+x 4) 1+x 4) >0 then when x0 is easy to know and only needs to consider x>0, find the maximum value of the original formula.

then when x>0, (1+x 2+x 4) 1+x 4)] x[ (1+x 2+x 4) Brighten (1+x 4)]*1+x 2+x 4)+ 1+x 4)] x*[ 1+x 2+x 4)+ 1+x 4)]}

x 2) {x*[ 1+x 2+x 4)+ 1+x 4)]}x [ 1+x 2+x 4)+ 1+x 4)] From the mean theorem, it can be seen that

x (-2) +x 2 2 [x (-2) *x 2] = 2 (equal sign if and only if x (-2) x 2, i.e. x 1).

So when x=1, [ 1+x 2+x 4 1+x 4)] x has a maximum value of 1 [ 2+1) +2]= 3 2

Answer: f(x) = (x -4x+13) + x -2x+3) = [(x-2) +3 ]+x-1) +2) ]f(x) represents the sum of the distances between points (x,0) on x to points (2,3) and points (1,- 2).

When three points are in a straight line, the sum of the distances is the minimum distance between the above two fixed points, so: f(x)>= [(2-1) +3+ 2) ]= (1+9+6 2+2).

So: the minimum value is (12+6 2).

Hello!!!

Do it with a geometric approach:

f(x) = (x 2-2x+2) + x 2-4x+8)) = ((x-1) 2+1) + x-2) 2+4) The equivalence of the problem is the minimum distance from x(x,0) to points a(1,1) and b(2,2).

Draw it in a planar Cartesian coordinate system and find the symmetry point A of a'(1,-1), there is symmetry to know, |a'b|The distance is sought.

The answer is root number 10

If there's anything you don't understand, ask me again.

Thank you!!!

sqrt[(x^2-2)^2+(x-3)^2]-sqrt[(x^2-1)^2+x^2]

The original problem is transformed into finding a point p on the parabola x=y 2 such that its distance from a(2,3) minus the distance from b(1,0) is maximized.

Drawing the diagram and using the triangular inequality pa-pb<=ab, we can see that the maximum value is ab=sqrt(1+9)=root number(10).

If it's not too troublesome, just ask for it.

f(x)=root[(x 2-1) 2+(x-3) 2]-root[(x 2-2) 2+(x) 2], so that y=x 2, then.

f(x)=root number[(y-1) 2+(x-3) 2]-root number[(y-2) 2+(x) 2] Its physical meaning is the difference between the distance between the two points on the function y=x 2 from the point (x,y) to (3,1),(0,2) respectively, find its maximum.

In the left side of the y-axis, the difference between the distance from the two points to the two points must be the largest, because the straight line between the two points is the shortest, so the maximum value is (3,1),(0,2) the distance between the two points, that is, the root number [(3-0) 2+(1-2) 2] = the root number 10

Solution: 2x 0 5-3x+4 0, x 0 5-2x 0 can be deduced from the domain of the function as (- 0] [2,+ When x 0, y= (2x 0 5-3x+4) is the subtraction function, y= (x 0 bending beam 5-2x) is also the subtraction function f(x) = 2x 0 5-3x+4) + x burn date0 5-2x) is the subtraction function When x 2, y= (2x 0 5-3x+4) is the increment function buried in the section, y= (x 0 5-2x) is also the increment function f(x) is the increment function In summary, fmin(x)=min=min=2 The minimum value of f(x) is 2