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dainel40

  • 3 years ago

f(x)=sqrt(5-e^(2x))

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  1. dainel40
    • 3 years ago
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    I need the domain

  2. anonymous
    • 3 years ago
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    make sure \[5-e^{2x}\geq 0\]

  3. dainel40
    • 3 years ago
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    ok but im stuck there

  4. anonymous
    • 3 years ago
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    \[5\geq e^{2x}\] \[\ln(5)\geq 2x\] \[\frac{\ln(5)}{2}\geq x\]

  5. dainel40
    • 3 years ago
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    so the domain would be

  6. dainel40
    • 3 years ago
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    I don't know how to write the domain for it because I keep getting it wrong

  7. anonymous
    • 3 years ago
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    maybe they want a decimal?

  8. dainel40
    • 3 years ago
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    How can I write it they give me this format ([ ]) U ([ ])

  9. some_someone
    • 3 years ago
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    f(x)=~(5-e^(2x)) The domain of an expression is all real numbers except for the regions where the expression is undefined. This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0. All of these are undefined and therefore are not part of the domain. (5-e^(2x))<0 Solve the equation to find where the original expression is undefined. x>(ln(5))/(2)_x<Z>APPR<z>0.8047 The domain of the rational expression is all real numbers except where the expression is undefined. x<=(ln(5))/(2)_(-<Z>I<z>,(ln(5))/(2)]

  10. some_someone
    • 3 years ago
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    \[x<=(\ln(5))/(2)_(-<Z>I<z>,(\ln(5))/(2)]\]

  11. dainel40
    • 3 years ago
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    ok thanks

  12. dainel40
    • 3 years ago
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    how can i find the inverse of that same question?

  13. dainel40
    • 3 years ago
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    i get a big equation that doesnt make any sense

  14. some_someone
    • 3 years ago
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    Inverse: f(x)=~(5-e^(2x)) To find the inverse of the function, interchange the variables and solve for f^(-1)(x). x=~(5-e^(2f)-1^(x)) Since f^(-1)(x) is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. ~(5-e^(2f)-1^(x))=x To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(5-e^(2f)-1^(x)))^(2)=(x)^(2) Simplify the left-hand side of the equation. 5-e^(2f)-1^(x)=(x)^(2) Expand the exponent (2) to the expression. 5-e^(2f)-1^(x)=(x^(2)) Remove the parentheses around the expression x^(2). 5-e^(2f)-1^(x)=x^(2) Since 5 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 5 from both sides. -e^(2f)-1^(x)=-5+x^(2) Reorder the polynomial -5+x^(2) alphabetically from left to right, starting with the highest order term. -e^(2f)-1^(x)=x^(2)-5 Multiply each term in the equation by -1. -e^(2f)-1^(x)*-1=x^(2)*-1-5*-1 Multiply -e^(2f^(-1)(x)) by -1 to get e^(2f^(-1)(x)). e^(2f)-1^(x)=x^(2)*-1-5*-1 Multiply x^(2) by -1 to get -x^(2). e^(2f)-1^(x)=-x^(2)-5*-1 Multiply -5 by -1 to get 5. e^(2f)-1^(x)=-x^(2)+5 Take the natural logarithm of both sides of the equation to remove the variable from the exponent. ln(e^(2f)-1^(x))=ln(-x^(2)+5) The natural logarithm of e^(2f^(-1)(x)) is 2f^(-1)(x). (2f^(-1)(x))=ln(-x^(2)+5) Remove the parentheses around the expression 2f^(-1)(x). 2f^(-1)(x)=ln(-x^(2)+5) Divide each term in the equation by 2. (2f^(-1)(x))/(2)=(ln(-x^(2)+5))/(2) Cancel the common factor of 2 in (2f^(-1)(x))/(2). (<X>2<x>f^(-1)(x))/(<X>2<x>)=(ln(-x^(2)+5))/(2) Remove the common factors that were cancelled out. f^(-1)(x)=(ln(-x^(2)+5))/(2)

  15. dainel40
    • 3 years ago
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    ok thanks

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