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Night reader wrote:If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =
A. 0.5
B. 1
C. 3
D. 4
E. 6
thanks Anshu, cool solutionanshumishra wrote:Night reader wrote:If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =
A. 0.5
B. 1
C. 3
D. 4
E. 6
f(3k) = f(k1)
=> 5-6k = 5-2(k+1)
=> k = 1/2
f(k) = 5 - 2k = 4
D
It is recommended to be careful while using this method to solve these kind of function problem. This is applicable here because it the given function is a linear one. If it is not, for example say f(x) = x², then this method may not give correct results or may miss some possible results.Night reader wrote:i did a bit diff. 3k=k+1, k=1/2 => f(1/2)=5-2*1/2=4
Anurag we can always use differentialsAnurag@Gurome wrote:It is recommended to be careful while using this method to solve these kind of function problem. This is applicable here because it the given function is a linear one. If it is not, for example say f(x) = x², then this method may not give correct results or may miss some possible results.Night reader wrote:i did a bit diff. 3k=k+1, k=1/2 => f(1/2)=5-2*1/2=4
Anshu's solution is full proof and applicable in general.
y=x^n => y`=(x^n)` => y=n*x^(n-1) and so on; I know what is functional relationship and lim-s of y with different functions vary...
We are given that f(x) = 5 - 2x. So f(3k) = 5 - 2(3k) = 5 - 6k and f(k + 1) = 5 - 2(k + 1) = 3 - 2k. Since f(3k) = f(k + 1), we equate the two expressions, and thus:Night reader wrote:If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =
A. 0.5
B. 1
C. 3
D. 4
E. 6
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