Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

satellite73Best ResponseYou've already chosen the best response.0
make sure \[5e^{2x}\geq 0\]
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
ok but im stuck there
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
\[5\geq e^{2x}\] \[\ln(5)\geq 2x\] \[\frac{\ln(5)}{2}\geq x\]
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
so the domain would be
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
I don't know how to write the domain for it because I keep getting it wrong
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
maybe they want a decimal?
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
How can I write it they give me this format ([ ]) U ([ ])
 one year ago

some_someoneBest ResponseYou've already chosen the best response.0
f(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)]
 one year ago

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

dainel40Best ResponseYou've already chosen the best response.0
how can i find the inverse of that same question?
 one year ago

dainel40Best ResponseYou've already chosen the best response.0
i get a big equation that doesnt make any sense
 one year ago

some_someoneBest ResponseYou've already chosen the best response.0
Inverse: 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)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.