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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0make sure \[5e^{2x}\geq 0\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0\[5\geq e^{2x}\] \[\ln(5)\geq 2x\] \[\frac{\ln(5)}{2}\geq x\]

dainel40
 2 years ago
Best ResponseYou've already chosen the best response.0so the domain would be

dainel40
 2 years ago
Best ResponseYou've already chosen the best response.0I don't know how to write the domain for it because I keep getting it wrong

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0maybe they want a decimal?

dainel40
 2 years ago
Best ResponseYou've already chosen the best response.0How can I write it they give me this format ([ ]) U ([ ])

some_someone
 2 years ago
Best ResponseYou've already chosen the best response.0f(x)=~(5e^(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. (5e^(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)]

some_someone
 2 years ago
Best ResponseYou've already chosen the best response.0\[x<=(\ln(5))/(2)_(<Z>I<z>,(\ln(5))/(2)]\]

dainel40
 2 years ago
Best ResponseYou've already chosen the best response.0how can i find the inverse of that same question?

dainel40
 2 years ago
Best ResponseYou've already chosen the best response.0i get a big equation that doesnt make any sense

some_someone
 2 years ago
Best ResponseYou've already chosen the best response.0Inverse: f(x)=~(5e^(2x)) To find the inverse of the function, interchange the variables and solve for f^(1)(x). x=~(5e^(2f)1^(x)) Since f^(1)(x) is on the righthand side of the equation, switch the sides so it is on the lefthand side of the equation. ~(5e^(2f)1^(x))=x To remove the radical on the lefthand side of the equation, square both sides of the equation. (~(5e^(2f)1^(x)))^(2)=(x)^(2) Simplify the lefthand side of the equation. 5e^(2f)1^(x)=(x)^(2) Expand the exponent (2) to the expression. 5e^(2f)1^(x)=(x^(2)) Remove the parentheses around the expression x^(2). 5e^(2f)1^(x)=x^(2) Since 5 does not contain the variable to solve for, move it to the righthand 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)*15*1 Multiply e^(2f^(1)(x)) by 1 to get e^(2f^(1)(x)). e^(2f)1^(x)=x^(2)*15*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)
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