If it is further given that \(f(x) = -x^3 + x^2 + x + 2\): Determine the coordinates of the turning points of \(f\).

Shifting \(f\) by \(k\) units vertically moves \(A\) to \(y = 9 + k\), \(B\) to \(y = 8 + k\) and \(C\) to \(y = 7 + k\).

For three \(x\)-intercepts the maximum must be above the \(x\)-axis and the minimum below it:

\[9 + k \gt 0 \text{ and } 8 + k \lt 0\]

For all three to be positive the \(y\)-intercept must be below the \(x\)-axis: \(7 + k \lt 0\) (already true when \(8 + k \lt 0\)).

\(-9 \lt k \lt -8\)

\[g(x) = -2x^{\frac{1}{2}}(x^2 - 2x + 1)\]\[g(x) = -2x^{\frac{5}{2}} + 4x^{\frac{3}{2}} - 2x^{\frac{1}{2}}\]\[g'(x) = -2\left(\frac{5}{2}\right)x^{\frac{3}{2}} + 4\left(\frac{3}{2}\right)x^{\frac{1}{2}} - 2\left(\frac{1}{2}\right)x^{-\frac{1}{2}}\]

\(g'(x) = -5x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - x^{-\frac{1}{2}}\)

\[f(x + h) = -6(x + h)^2 = -6x^2 - 12xh - 6h^2\]\[f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\]\[= \lim_{h \to 0}\frac{-6x^2 - 12xh - 6h^2 + 6x^2}{h}\]\[= \lim_{h \to 0}\frac{-12xh - 6h^2}{h}\]\[= \lim_{h \to 0}\frac{h(-12x - 6h)}{h}\]\[= \lim_{h \to 0}(-12x - 6h)\]

\(f'(x) = -12x\)

\[f'(x) = 3x^2 - 2x - 5 = 0\]\[(x + 1)(3x - 5) = 0\]

\(x = -1\) (this is M) or \(x = \frac{5}{3}\)

\[f\left(\frac{5}{3}\right) = \left(\frac{5}{3}\right)^3 - \left(\frac{5}{3}\right)^2 - 5\left(\frac{5}{3}\right) - 3 = -\frac{256}{27}\]

\(N\left(\frac{5}{3}\,;\,-\frac{256}{27}\right)\) or \(N(1{,}67\,;\,-9{,}48)\)

Divide \(f\) by \((x - 2)\):

\[f(x) = (x - 2)(x^2 - 6x - 7) = (x - 2)(x - 7)(x + 1)\]

\(x\)-intercepts: \(A(-1\,;\,0)\), \(B(2\,;\,0)\), \(C(7\,;\,0)\)

The product is negative when \(f\) and \(f''\) have opposite signs:

\(f''(x) \lt 0\) for \(x \lt \frac{8}{3}\) and \(f(x) \gt 0\) for \(-1 \lt x \lt 2\)

\(f''(x) \gt 0\) for \(x \gt \frac{8}{3}\) and \(f(x) \lt 0\) for \(2 \lt x \lt 7\)

\(-1 \lt x \lt 2\) or \(\frac{8}{3} \lt x \lt 7\)

Write each term as a power of \(x\):

\[\frac{-6x^{\frac{1}{3}} + 2}{x^4} = -6x^{-\frac{11}{3}} + 2x^{-4}\]\[D_x = -6\left(-\frac{11}{3}\right)x^{-\frac{14}{3}} + 2(-4)x^{-5}\]

\(= 22x^{-\frac{14}{3}} - 8x^{-5}\)