Answer: 1a is odd with end behavior as x approaches negative infinity f(x) will also approach negative infinity, and as x approaches positive infinity f(x) will also approach positive infinity.
Step-by-step explanation:
For 1a:
1.) We determine if the function is even or odd. A function is even if f(−x)=f(x). A function is odd if f(−x)=−f(x). A function is neither if neither of the conditions above are satisfied.
[tex]f(x)=x^3+x\\f(-x)=(-x)^3+(-x)\\f(-x)=-x^3-x\\\\-x^3-x=-(x^3+x)[/tex]
So this function is odd. Function even or odd can also be found graphically: if a function is even, the graph is symmetrical about the y-axis; if the function is odd, the graph is symmetrical about the origin.
2.) Find end behavior. The "arms" of even functions point in the same direction, while the "arms" of odd functions go in opposite directions. Look at the leading coefficient to see if it's positive or negative. It's 1, so positive. In this case, the left "arm" will point down and the right "arm" will point up, like the graph of [tex]y=x^3[/tex].
3.) We find the zeroes by factoring.
[tex]f(x)=x^3+x\\f(x)=x(x^2+1)[/tex]
Our zeroes are 0 and i. i isn't real, so we discard it.
4.) We find the multiplicity our zero, or simply how many times it's multiplied in the function. 0 has a multiplicity of 1. On the graph, zeroes with a multiplicity of 1 go straight through, zeroes with a multiplicity of 2 bounce right off, and zeroes with a multiplicity of 3 stop and stay for a while.
5.) Graphing time! Graph the zeroes first on the y-axis, which would be (0, 0) in our case. Draw arms on the zeroes that point in the correct directions as stated above. Fill in the in betweens (even though there's nothing in between here) by following the multiplicity. The graph would go straight through at (0, 0). Your finished graph should look something like [tex]y=x^3[/tex] but with a smoother curve at the middle.
For 1b, just repeat the steps. Comment or message me if you need any further help!
Also nice pun lol
