Answer: (a) \(f(x)=\pm3(x+2)(x-1)(x-4)\); (b) \(\frac19\leq x\leq3\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

(a) The graph of \(\lvert f(x)\rvert\) touches the \(x\)-axis at

\(x=-2,\quad x=1,\quad x=4.\)

So the cubic \(f(x)\) has factors

\((x+2),\quad (x-1),\quad (x-4).\)

Therefore

\(f(x)=k(x+2)(x-1)(x-4).\)

From the graph, \(\lvert f(0)\rvert=24\). Now

\((0+2)(0-1)(0-4)=8.\)

Hence \(\lvert 8k\rvert=24\), so \(\lvert k\rvert=3\).

Thus the possible expressions are

\(f(x)=3(x+2)(x-1)(x-4)\)

or

\(f(x)=-3(x+2)(x-1)(x-4).\)

(b) Since both sides are non-negative, square both sides:

\((5x-2)^2\leq(4x+1)^2.\)

Expand and simplify:

\(25x^2-20x+4\leq16x^2+8x+1.\)

So

\(9x^2-28x+3\leq0.\)

Factorise:

\((9x-1)(x-3)\leq0.\)

The quadratic is less than or equal to zero between its roots, so

\(\frac19\leq x\leq3.\)