The function f is defined as f(x)=(2x+1)(x-3) where x≠q3. Find the value of k so that the inverse of f is f^(-1)(x)=(3x+1)(x-k) 相关知识点: 试题来源: 解析 The correct answer is (2). y=f(x)=(2x+1)(x-3) where x≠q3. Interchange x and y.x=(2y+1...
Answer to: Find the values of k so that the function f(x) = \left\{\begin{matrix} 2k^2x + 1 & x \gt 1 \\ k- 2 & x \lt 1 \end{matrix}\right. ...
D. The value of k cannot be determined. Continuous Function: A function {eq}f(x) {/eq} is said to be continuous at {eq}x= a {/eq}, if and only if and only if the limit of the function at this value and the value of the functio...
I am working on a unit step function, and what I want to do is to find the maximum value of a constant that keeps my function Y value as close as 1.3 as possible. This is the function of the code: ThemeCopy functionH = transferH(K) ...
Find the maximum value of K such that it is possible to find the sub-sequences A and B which satisfies the above conditions. Constraints: 0 < N <= 10^5Things which I observed are: The value of K = 0 if the number of characters in the given string are all ...
So I'd like to find the maximum value of the column Return in df when changing the variables ev,bv,vc & dv. I'd like them to be between in the intervals of ev: 0.2-4, bv: 300-600, vc: 0-1000 & dv: 0-1000. When running my code it seem like the function stops at ...
m1]. This is easily seen that if the left point of the function is smaller than the right, then either of these two points are in the area of "lifting" function, or only the left point is there. In any case, this means that the peak is meaningful to search on only the ...
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Find the value of constant K that makes the function below continuous at x=5. f(x)={x2−5xx−5 if x<5Kx−3 if x≥5 Continuous Function: A piecewise function is said to be continuous at a point when the on...
Find the value of {eq}k {/eq} so that {eq}f {/eq} is continuous at {eq}x = 1 {/eq}. {eq}\displaystyle f (x) = \begin{cases}3 k x^2 + 2 x - 4, & x <1\\ 7 k \sin \bigg(\dfrac {\...