1. Find the maximum value of the function$$ y = \frac { 2 6 \sqrt { 3 } } { 3 } \cos x + \frac { 1 3 \sqrt { 3 } } { 3 } x - \frac { 1 3 \sqrt { 3 } \pi } { 1 8 } + 1 1 $$on the interval [0, $$ \frac { \pi } { 2
For the function f(x) = sin(x) + cos(x), the maximum value of the function in the interval [0, 2π] is: A. √2 B. 1 C. 2 D. -√2 相关知识点: 试题来源: 解析 A。f(x) = √2sin(x + π/4),在区间 [0, 2π] 中,最大值为 √2。
The maximum value of the function ƒ(x) = e^x + x ln x on the interval 1 le x le 2 is
The maximum value of the function f ( x ) = 3 x 3 - 18 x 2 + 27 x - 40 on the set S = { x ∈ R : x 2 + 30 ≤ 11 x } is: A - 122 B - 222 C 222 D 122 Video Solution Struggling with Differential Ca... ? Get free crash course ...
Let f(x) be a function defined in a domain D then the maximum value of the function will attain either at the critical points of the function or at the boundary points of the interval. The maximum of all the value is called the...
解析 Let .Now, will be negative when is positive i.e., when and are both positive. Also, we know that and both are positive in the first quadrant. Then, will be negative when Thus, we consider By second derivative test, will be the maximum at and the maximum value of is ...
The maximum value of the function f(x) is s ; the minimum value of the function f(x) is 相关知识点: 试题来源: 解析 13. Let the function f(x)=-x^2+4x ,1x4. The maximum value of the function f(x) is 4_; the minimum value of the function f(x) is _ 0 ...
To do this, we have to clear one of the variables from the constraint and substitute it in the function to be optimized.Answer and Explanation: In order find the maximum value of the function {eq}f\left( {x,y,z} \right) = x + 2y ...
The sign for the second derivative is negative. So, the point of maxima is x=0 Next, substitute x = 0 in the first equation to obtain the maximum value of the function Y = 0 – 0+ 6 Y = 6 Types of Maximum Function There are two types of maxima functions that exist ...
If we have to find a maximum or minimum value of any function, product of two variables in this case, under a condition, straight line in this case, we apply Lagrange's multipliers. We find the gradient of the function and the condition and introduce one paramet...