Sketching the graph of a cubic

If $f'(x)$ changes from $+$ to $-$ around a point where $f'(x)=0$, there is a local maximum; from $-$ to $+$, a local minimum. Substitute into the original function to get the value. For example, $y=x^3-12x$ has $y'=3x^2-12=3(x+2)(x-2)$, so it has a local maximum $16$ at $x=-2$ and a local minimum $-16$ at $x=2$.

As $x$ grows without bound, the behavior of $y$ depends on the coefficient of $x^3$. If it is positive, $y$ grows without bound; if negative, $y$ decreases without bound.

If $f'(x)$ never changes sign, there is no local extremum. $f'(x)=3x^2+4$ is always positive, so $f(x)$ keeps increasing and has no local extremum.

Examples

1. 01For which $f'(x)$ below does $f(x)$ have no local extremum?$\displaystyle$A$\displaystyle 3x^{2}-3x$B$\displaystyle 3x^{2}-1$C$\displaystyle 3x^{2}-3$D$\displaystyle 3x^{2}+4$AnswerHide

   Answer$\displaystyle 3x^{2}+4$

   Steps

   1. $\displaystyle 3x^2$$\displaystyle \ge 0$
   2. Adding a positive constant
   3. $\displaystyle 3x^2+4$$\displaystyle \ge 4 > 0$
2. 02This cubic has a local maximum at x=-2. What is the local maximum value?$\displaystyle y=x^3-12x$AnswerHide

   Answer$\displaystyle 16$

   Steps

   1. $\displaystyle (-2)^3 - 12 \cdot (-2)$$\displaystyle = -8 + 24$$\displaystyle = 16$
3. 03This cubic has a local minimum at x=3. What is the local minimum value?$\displaystyle y=x^3-27x$AnswerHide

   Answer$\displaystyle -54$

   Steps

   1. $\displaystyle (3)^3 - 27 \cdot 3$$\displaystyle = 27 - 81$$\displaystyle = -54$
4. 04This cubic has a local maximum at x=-1. What is the local maximum value?$\displaystyle y=x^3-3x$AnswerHide

   Answer$\displaystyle 2$

   Steps

   1. $\displaystyle (-1)^3 - 3 \cdot (-1)$$\displaystyle = -1 + 3$$\displaystyle = 2$

FAQ

Q1If $f'(x)=0$ somewhere, does $f$ have a local extremum there?AnswerHide

A

Not necessarily. For $f(x)=x^3$, $f'(0)=0$, but $f'(x)=3x^2\ge 0$ never changes sign, so there is no extremum. A cubic has local extrema exactly when the quadratic $f'(x)$ has a positive discriminant, so that $f'(x)=0$ has two distinct real solutions.

Q2For $y=-x^3+6x$, what happens to $y$ as $x$ grows?AnswerHide

A

It decreases without bound. Writing $-x^3+6x=x^3\left(-1+\frac{6}{x^2}\right)$, the factor in parentheses approaches $-1$, so $y$ becomes a larger and larger negative number.