Why the graph of $f(x)=x^4$ has no inflection points

Key concept: A clean way to test inflection points is to check where concavity changes. For $f(x)=x^4$, the second derivative gives the whole story, and the sign never flips.

Step by step

Step 1: Differentiate twice

Start with

$f(x)=x^4$

First derivative:

$f'(x)=4x^3$

Second derivative:

$f''(x)=12x^2$

Step 2: Find where concavity could change

Set the second derivative equal to zero:

$12x^2=0 \Rightarrow x=0$

That is the only candidate.

Step 3: Check the sign of $f''(x)$ on both sides

For every real $x$,

$12x^2 \ge 0$

and in fact $f''(x)>0$ whenever $x\neq 0$.

So on both sides of $x=0$, the graph remains concave up.

Step 4: Decide whether an inflection point exists

An inflection point needs a change in concavity. Here, there is no change:

• left of $0$: concave up
• right of $0$: concave up

So $x=0$ is not an inflection point.

Conclusion

$\boxed{f(x)=x^4\text{ has no points of inflection}}$

Pitfall alert

A common mistake is to think that because $f''(0)=0$, there must be an inflection point there. That is not enough. You still have to check whether $f''$ changes sign. If it stays positive or stays negative, there is no inflection point.

Try different conditions

If the function were $f(x)=x^3$, then $f''(x)=6x$, which does change sign at $x=0$. That is exactly the kind of situation where an inflection point does occur. The difference is not whether $f''$ is zero, but whether concavity switches.

Further reading

second derivative test, concavity, inflection point