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2 years ago

F(x)=2x and g(x)=square root of x what is the domain of f(g(x))?

Mathematics

1 answer:

Dmitriy789 [7]2 years ago

5 0

Answer:

$x\ge 0$

Step-by-step explanation:

The given functions are:

$f(x)=2x$

and

$g(x)=\sqrt{x}$

Now,

$f(g(x))=f(\sqrt{x} )$

$f(g(x))=2\sqrt{x}$

The domain refers to values of $x$ for which $f(g(x))=2\sqrt{x}$ is defined.

The expression under the radical sign must not be negative.

The domain is

$x\ge 0$

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Which of the following graphs represents the function f(x) = x4 - 2x3 - 3x2 + 4x + 1?

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Answer: Graph D will be correct graph for the given function.

Explanation:

Given function $f(x) = x^4-2x^3-3x^2+4x+1$

Since it is a bi-quadratic equation thus it must have 4 roots and (0,1) is one of its point.

Moreover, the degree of the function is even thus the end behavior of the function is $f(x)\to+\infty$ , as $x\to-\infty$ and $f(x)\to+\infty$ as $x\to+\infty$

In graph A, function has four root but it does not have the end behavior same as function f(x).( because in this graph $f(x)\to-\infty$ , as $x\to-\infty$ and $f(x)\to-\infty$ , as $x\to+\infty$ .) so, it can not be the graph of given function.

In graph B, neither it has four root nor it has the end behavior same as function f(x).(because in this graph $f(x)\to+\infty$ as $x\to-\infty$ and $f(x)\to-\infty$ as $x\to+\infty$ .) so, it can not be the graph of given function.

In graph C, neither it has four root nor it has the same end behavior as function f(x).(because in this graph $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to+\infty$ as $x\to\infty$ .) so, it also can not be the graph of given function.