The function [tex]f(x)=5cos(x)+1[/tex] is the cosine curve with 5 as the amplitude and is shifted 1 units up.
Let's check all the 4 choices:
A.
The period of a function in the form [tex]f(x)=Acos(x)+B[/tex] (which is the function in this problem) has period [tex]2\pi[/tex]. Period would have changed if there was any coefficient of x in the argument. But it doesn't so period is NOT [tex]10\pi[/tex].
B.
The amplitude of a function in the form [tex]f(x)=Acos(x)+B[/tex] (which is the function in this problem) is A. But A is 5 here so the amplitude is definitely NOT 2.5.
C.
A zero of the function is given as [tex](\frac{\pi}{2},0)[/tex], which means that if we plug in [tex]\frac{\pi}{2}[/tex] into x, it should give us 0. Let's check:
[tex]5cos(\frac{\pi}{2})+1\\=5(0)+1\\=1[/tex]. It's not true.
D.
We know that this function has an amplitude of 5, so the part [tex]5cos(x)[/tex] ranges from -5 to 5. But this part is shifted 1 units up, so the max amplitude is [tex]5+1=6[/tex] and the min is [tex]-5+1=-4[/tex]. So we can say that the range of the function is the set of real numbers from -4 to 6. This choice is correct.
ANSWER: Last choice, choice D.
The statement which is true for f(x) = 5cos(x) +1 is; Choice D; The range of the function is the set of real numbers -4≤ y ≤ 6
From comparison with equation of the form;
where;A = amplitude
We can conclude that this function has an amplitude of 5, so the portion 5cos(x) of the function ranges from -5 to 5.
However, this part is translated 1 units up as denoted by the +1.
Ultimately,
Therefore, we can conclude that the range of the function is the set of real numbers from -4 to 6.
Read more on Range of a function;
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