The function f(x)=x^3-6x^2+9x+25 has

The correct answer is Option (1) → A local maxima at \( x=1 \) and a local minima at \( x=3 \).

The critical points come from f′(x) = 0, and the second derivative test decides their nature.

    \( \displaystyle f'(x)=3x^2-12x+9 \)

    \( \displaystyle =3(x-1)(x-3) \)

    \( \displaystyle f'(x)=0\ \Rightarrow\ x=1\ \text{or}\ x=3 \)

    \( \displaystyle f''(x)=6x-12 \)

    \( \displaystyle f''(1)=-6\lt 0:\ \text{local maximum at}\ x=1 \)
    \( \displaystyle f''(3)=6\gt 0:\ \text{local minimum at}\ x=3 \)

Tip: A cubic with a positive leading coefficient and two critical points always has its local maximum at the smaller one and its local minimum at the larger one.