Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 07/Solution 1

(i) Incorrect. $f(x)=x^{3},\ f'(x)=3x^{2},\ f'(0)$ exists and $f'(0)=0$, but the derivative is everywhere non-negative so $x=0$ is NOT an extremum.

(ii) Incorrect. $f(x)=|x|$ has a minimum at $x=0$, but the derivative does NOT exist at $x=0$ (a corner).

(iii) Incorrect. A function can be decreasing at (-1) and increasing at 1, but have a local max at $x=0$. Consider a function $f'(x)=x(x-{\frac {1}{2}})(x+{\frac {1}{2}})$.

(iv) Correct. Because if $f''(0)=0$ and $f'''(0)\neq 0$, this means that $(0,0)$ is an extremum for $f'$. (Applying the second derivative test for $f'$). Now suppose for example $f'''(0)>0$ which implies that $(0,0)$ is a minimum point for $f'$, and therefore by the definition for any $x$ in the neighborhood of 0 we have $f'(x)>0$, which is equivalent to $f$ being increasing in the neighborhood of 0, so $f$ cannot have an extremum at $x=0$ . By the similar argument, we can get the same result in the case of $f'''(0)<0$.