Question

How do you sketch the general shape of `f(x)=-x^5+3x^3+2` using end behavior?

Answer 1

See explanation...

Explanation:

End behaviour will help, but it's not really enough...

The end behaviour is determined purely by the term of highest degree, that is the term `-x^5`.

Since this term is of odd degree with negative coefficient, we find that:

`lim_(x->oo) f(x) = -oo`

`lim_(x->-oo) f(x) = oo`

If `f(x)` lacked this term, then its end behaviour would be determined by the term of next largest degree, `3x^3`.

Since this term is of odd degree with positive coefficient, we have:

`lim_(x->oo) 3x^3+2 = oo`

`lim_(x->-oo) 3x^3+2 = -oo`

... precisely the opposite of `f(x)`.

So one question we might ask is whether the term in `x^3` has a large enough coefficient to cause a kink in the curve for small values of `x`.

The answer is yes. Since there is only a constant term following, the `3x^3` term will dominate the `-x^5` term for values of `x` in `(-sqrt(3), sqrt(3))`.

Next note that the constant term `2` will result in a `y` intercept at `(0, 2)`.

Next note that `f(-1) = 1-3+2 = 0`, so there is an `x` intercept at `(-1, 0)`.

We can supplement our analysis by looking at the derivative:

`f'(x) = -5x^4+9x^2 = -5x^2(x^2-9/5)`

Hence there is a local minimum at `x = -sqrt(9/5) = -(3sqrt(5))/5` a point of inflexion at `x=0` and a local maximum at `x = sqrt(9/5) = (3sqrt(5))/5`.

You can calculate a few example points too to help find that the curve looks something like this...

graph{-x^5+3x^3+2 [-10, 10, -5, 5]}