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liubo4ka [24]
3 years ago
5

Let f(x) =x+1 and g(x)=1/x. The graph of (f o g)(x) is shown below. What is the range of (f o g)(x)?

Mathematics
2 answers:
sveta [45]3 years ago
6 0
Looks like all real numbers except 1. It touch's all the y except 1 aka range
Eddi Din [679]3 years ago
6 0

Answer:

all real numbers except y=1

Step-by-step explanation:

f(x)=x+1

g(x)=\frac{1}{x}

(f og)(x)= f(g(x))

Plug in 1/x for g(x)

(f og)(x)= f(g(x))=f(\frac{1}{x})

Now we plug in the fraction in f(x)

(f og)(x)= f(g(x))=f(\frac{1}{x})=\frac{1}{x}+1

Range of the function are the y values for which the function is defined

In the graph , we can see that the graph goes close to y=1 but does not cross y=1

So range is all real numbers except y=1

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Step-by-step explanation:

We are given the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}

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And we have our final answer.

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