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

Jun 14, 2017

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:

${\lim}_{x \to \infty} f \left(x\right) = - \infty$

${\lim}_{x \to - \infty} f \left(x\right) = \infty$

If $f \left(x\right)$ lacked this term, then its end behaviour would be determined by the term of next largest degree, $3 {x}^{3}$.

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

${\lim}_{x \to \infty} 3 {x}^{3} + 2 = \infty$

${\lim}_{x \to - \infty} 3 {x}^{3} + 2 = - \infty$

... precisely the opposite of $f \left(x\right)$.

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 $3 {x}^{3}$ term will dominate the $- {x}^{5}$ term for values of $x$ in $\left(- \sqrt{3} , \sqrt{3}\right)$.

Next note that the constant term $2$ will result in a $y$ intercept at $\left(0 , 2\right)$.

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

We can supplement our analysis by looking at the derivative:

$f ' \left(x\right) = - 5 {x}^{4} + 9 {x}^{2} = - 5 {x}^{2} \left({x}^{2} - \frac{9}{5}\right)$

Hence there is a local minimum at $x = - \sqrt{\frac{9}{5}} = - \frac{3 \sqrt{5}}{5}$ a point of inflexion at $x = 0$ and a local maximum at $x = \sqrt{\frac{9}{5}} = \frac{3 \sqrt{5}}{5}$.

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]}