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

1 Answer
Jun 14, 2017

Answer:

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]}