Can someone answer this problem, and explain it to me. Thank you! a) f(x) = e^(-x^2) b) f(x) = 1 + sin^-1(-x) c) f(x) = abs value e^x d) f(x) = e^(ln abs value (x) +1 Thanks so much!
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f(-1)=f(1) Even f(-1)=-f(1) Odd ex f(x)=X^2 is even
Originally Posted by calculuskid1 f(-1)=f(1) Even f(-1)=-f(1) Odd ex f(x)=X^2 is even There should be x not 1 or -1.
Still don't understand that is at all...
Originally Posted by KarlosK Still don't understand that is at all... if $\displaystyle \color{blue}f(-x)=f(x)$ then $\displaystyle \color{blue}f$ is an even function. if $\displaystyle \color{blue}f(-x)=-f(x)$ then $\displaystyle \color{blue}f$ is an odd function.
In other words, use the definition of "even" and "odd" function. $\displaystyle f(x)= e^{x^2}$ so $\displaystyle f(-x)= e^{(-x)^2}$ but $\displaystyle (-x)^2= x^2$ so $\displaystyle f(-x)= e^{x^2}= f(x)$.
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