Additional Mathematics – Equations, inequalities and graphs | e-Consult

Answer 3

Rewrite as \((x-2)(x+3)(x-4)+6 \le 0\). Let \(h(x)= (x-2)(x+3)(x-4)+6\).

Finding the zeros of \(h(x)\) (by trial or calculator) gives \(x\approx -3.41,\;2.00,\;4.41\).

Sign chart for \(h(x)\):

• \((-\infty,-3.41):\;h(x)>0\)
• \((-3.41,2.00):\;h(x)\le0\)
• \((2.00,4.41):\;h(x)>0\)
• \((4.41,\infty):\;h(x)\le0\)

Including the points where \(h(x)=0\) (i.e., \(f(x)=-6\)), the solution of \(f(x)\le-6\) is:

\[[-3.41,2]\cup[4.41,\infty)\]