Find the maximum and minimum values of the function f x x 2x on 0,10

It depends on what kind of function it is; If it is a one-time function, then the values of the function at the beginning and end of the closed interval [a,b] are its minimum and maximum values, respectively; If it is a quadratic function, it will be discussed on a case-by-case basis: (1) when the opening is upward, there is a minimum value in the defined domain; If you give an interval range, you also need to see that the interval includes vertices and does not include vertices, including vertices, then the vertices are the minimum value of the function, excluding vertices is the post, if the interval is on the right side of the symmetry axis of the function, then the function value of the starting point is the minimum value, if the interval is on the left side of the symmetry axis of the function, then the function value of the end point is the minimum value; (2) When the opening is downward, there is a maximum value in the defined domain; If you are given an interval range, it also depends on whether the interval includes vertices; If vertices are included, then the ordinate of the vertices is the maximum value of the function, if the vertices are not included and the interval is to the left of the axis of symmetry, then the end point is the maximum value of the function, and the function value of the opposite starting point is the maximum value of the function;

There is also the method of finding the minimum value of the logarithmic function of the exponential function, which discusses the monotonicity of the function in the given defined domain; Then let's find the maximum value of the function.

From the problem, we can find the axis of symmetry x=1 of this function, and we can see that in the interval of 0,10, the function increases at 0,1 and decreases at 1,10, so the maximum value is x=1 substitution; f(x)=1, the minimum value is substituted x=10, and f(x)=-80 is obtained

f(x)=-x²+2x

x-1)^2+1

When x=1.

f(x)=1

The maximum value ymax = 1

When x = 10f (x ) = -80

The minimum ymin = -80

Obviously, at x [0,2], f(x) increases monotonically.

0≤x≤21≤x+1≤3

1/3≤1/(x+1)≤1

2≤-2/(x+1)≤-2/3.

2≤f(x)≤-2/3.

Therefore, the maximum value is -2 3;

The minimum value is -2

Maximum 1, no minimum.

f(x) 2x 4x 1 is a parabola with an opening pointing downward, and the maximum value is the vertex value, i.e. the value at x=1, and f(1)=1

There is no minimum.

Alternatively, the f(x) 2x 4x 1 recipe can be morphed to.

f(x)＝－2（x-1）²+1

So the maximum value is 1 and the minimum value does not exist.

Derive. f'(x)=3x²+12x-15

3(x-1)(x+5)

Order f'(x)=0

Solution. x=1 or x=-5

Because. f(1)=-10, f(0)=-2, excite f(2)=0 so that f(x) is on [0,2], the maximum simple lead grip value is f(2)=0, and the minimum value is f(1)=-10.

Obviously, when x [0,2], f(x) is an increasing difference.

0 x 21 x+1 Reputation Rotten Skin 3

1/3≤1/(x+1)≤1

2 Accolades -2 (x+1) -2 3

2≤f(x)≤-2/3.

Therefore, the maximum value is -2 3;

The minimum value is -2

First, find the definition domain of the number of stools in Hanxiang Mountain.

x +x+3 4 Respectfully 0

i.e. x - x-3 4 0

i.e. (x-3 2) (x+1 2) 0

1/2≤x≤3/2

Then find the range of g(x)=-x +x+3 4 in [-1 2,3 only manuscript 2].

g（x）=-x-1/2）²+1

When x [-1 2,3 2], its minimum value is 0, (which is mandatory, the domain of the function definition), and the maximum value is g(1 2)=1

f(x) = g(x) has a minimum value of 0 and a maximum value of 1

The image method or derivative method obtains: f(x)[0,1] increases, [1,10] decreases, and the maximum value is obtained as 1 at x=1

Let the derivative of f(x) be zero, i.e. f'(x)=-2x+2=0 is solved to obtain x=1 in the interval [0,10]. In the end, the answer is solved.

Share a solution. Obviously, when x=0, f(x)=0. Again, at x r, x +1 2x, at x≠0, f(x) x (2x)=1 2.

f(x) has a maximum value of 1 2.

Again, f(x) is an odd function, and with respect to origin symmetry, f(x) has a minimum value of -1 2.

FYI.

Is x 1 together, if yes, the numerator and denominator are divided by x at the same time, and 1 (x 1 x), the denominator is a function of the pair, it is easy to know that x 1 has a maximum value of 1 2, and when x tends to 0 or + there is a minimum value of 0