If the function \( f(x) \) defined by
\( f(x)=x^3-3x+100,(x\in\mathrm{R}) \), (Where \( \mathrm{R} \) is set of real numbers), then \( f(x) \) has
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Local maxima at \( x=-1 \) and local minima at \( x=1 \)
The critical points come from \( f'(x)=0 \), and the second derivative test decides their nature.
\( \displaystyle f'(x)=3x^2-3=3(x-1)(x+1) \)
\( \displaystyle f'(x)=0\ \Rightarrow\ x=1\ \text{or}\ x=-1 \)
\( \displaystyle f''(x)=6x \)
\( \displaystyle f''(-1)=-6\lt 0\ \Rightarrow\ \text{local maximum at}\ x=-1 \)
\( \displaystyle f''(1)=6\gt 0\ \Rightarrow\ \text{local minimum at}\ x=1 \)
Why the other options are incorrect
Option (1) Local maxima at \( x=1 \) and local minima at \( x=0 \) — Incorrect. \( f'(0)=-3\ne 0 \), so x = 0 is not a turning point at all.
Option (2) Local maxima at \( x=0 \) and local minima at \( x=-1 \) — Incorrect. \( f'(0)=-3\ne 0 \), so x = 0 cannot be a local maximum.
Option (3) Local maxima at \( x=1 \) and local minima at \( x=-1 \) — Incorrect. This reverses the test: \( f''(1)\gt 0 \) means a minimum at x = 1, not a maximum.