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Otrada [13]
2 years ago
8

Let F: R → R be defined by f(x) = x”. Show that f'is one-to-one and onto.​

Mathematics
1 answer:
Brrunno [24]2 years ago
3 0

Answer:

both

Step-by-step explanation:

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Yuki888 [10]

Answer:

Final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

Step-by-step explanation:

Given problem is \sqrt[3]{x}\cdot\sqrt[3]{x^2}.

Now we need to simplify this problem.

\sqrt[3]{x}\cdot\sqrt[3]{x^2}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}

Apply formula

\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}

so we get:

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x

Hence final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

7 0
3 years ago
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