For the function f(x)=x^3-2x^2-4x-5 , which of the following statemen…

The correct answer is Option (2) → The local maximum value of \( f(x) \) at \( x=-\displaystyle\frac{2}{3} \) is \( \displaystyle\frac{13}{27} \).

The critical points come from \( f'(x)=0 \).

    \( \displaystyle f'(x)=3x^{2}-4x-4 \)

    \( \displaystyle =(3x+2)(x-2) \)

So \( x=-\frac{2}{3} \) and \( x=2 \). By the second derivative test with \( f''(x)=6x-4 \),

    \( \displaystyle f''(2)=8\gt0 \)

    \( \displaystyle f''\left(-\frac{2}{3}\right)=-8\lt0 \)

So there is a local minimum at x = 2 and a local maximum at \( x=-\frac{2}{3} \). The values are

    \( \displaystyle f(2)=8-8-8-5 \)

    \( \displaystyle =-13 \)

    \( \displaystyle f\left(-\frac{2}{3}\right)=-\frac{8}{27}-\frac{8}{9}+\frac{8}{3}-5 \)

    \( \displaystyle =\frac{-8-24+72-135}{27} \)

    \( \displaystyle =-\frac{95}{27} \)

Each option is a statement, and the answer is the one that is not correct.

Option (1) The function \( f(x) \) has local minima at \( x=2 \) — Correct. \( f''(2)=8\gt0 \).

Option (2) The local maximum value of \( f(x) \) at \( x=-\frac{2}{3} \) is \( \frac{13}{27} \). — Incorrect. The local maximum value is \( f\left(-\frac{2}{3}\right)=-\frac{95}{27} \).

Option (3) The turning points of the given curve are \( (2,-13) \) and \( \left(-\frac{2}{3},-\frac{95}{27}\right) \) — Correct. \( f(2)=-13 \) and \( f\left(-\frac{2}{3}\right)=-\frac{95}{27} \).

Option (4) The function has local maxima at \( x=-\frac{2}{3} \) — Correct. \( f''\left(-\frac{2}{3}\right)=-8\lt0 \).

So Option (2) is the only statement that is not correct.