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Monica [59]
3 years ago
11

25. To produce the graph of the function y=0.5cos(0.5x), what transformations should be applied to the graph of the parent funct

ion y=cot(x)?
a. a horizontal compression to produce a period of pi/2 and a vertical compression
b. a horizontal compression to produce a period of pi/2 and a vertical stretch
c. a horizontal stretch to produce a period of 2pi and a vertical compression
d. a horizontal stretch to produce a period of 2pi and a vertical stretch
Mathematics
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer:

C. A horizontal stretch to produce a period of 2\pi and a vertical compression.

Step-by-step explanation:

We are given the parent function as y= \cot x

It is given that, transformations are applied to the parent function in order to obtain the function y=0.5\cot (0.5x) i.e. y=\frac{1}{2}\cot (\frac{x}{2})

That is, we see that,

The parent function y= \cot x is stretched horizontally by the factor of \frac{1}{2} which gives the function y=\cot (\frac{x}{2}).

Further, the function is compressed vertically by the factor of \frac{1}{2} which gives the function y=\frac{1}{2}\cot (\frac{x}{2}).

Now, we know,

If a function f(x) has period P, then the function cf(bx) will have period \frac{P}{|b|}.

Since, the period of y= \cot x is \pi, so the period of y=\frac{1}{2}\cot (\frac{x}{2}) is \frac{\pi}{1/2} = 2\pi

Hence, we get option C is correct.

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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
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Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

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This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

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