(a) To find \(ff(x)\), substitute \(f(x) = 2x^2 + 3\) into itself:

\(ff(x) = f(f(x)) = f(2x^2 + 3) = 2(2x^2 + 3)^2 + 3\).

Expand \((2x^2 + 3)^2\):

\((2x^2 + 3)^2 = 4x^4 + 12x^2 + 9\).

Substitute back:

\(ff(x) = 2(4x^4 + 12x^2 + 9) + 3 = 8x^4 + 24x^2 + 18 + 3 = 8x^4 + 24x^2 + 21\).

(b) Solve \(ff(x) = 34x^2 + 19\):

\(8x^4 + 24x^2 + 21 = 34x^2 + 19\).

Rearrange to form a polynomial equation:

\(8x^4 + 24x^2 + 21 - 34x^2 - 19 = 0\).

Simplify:

\(8x^4 - 10x^2 + 2 = 0\).

Factorize:

\(2(x^2 - 1)(4x^2 - 1) = 0\).

Solve each factor:

\(x^2 - 1 = 0\) gives \(x = 1\).

\(4x^2 - 1 = 0\) gives \(x = \frac{1}{2}\).

Thus, \(x = 1\) or \(x = \frac{1}{2}\).

(a) The function \(f(x) = (x - 2)^2 - 4\) is a quadratic function with vertex at \((2, -4)\). Since it opens upwards, the minimum value is \(-4\). Thus, the range is \(f(x) > -4\).

(b) To find \(f^{-1}(x)\), set \(y = (x - 2)^2 - 4\) and solve for \(x\):

\(y + 4 = (x - 2)^2\)

\(x - 2 = \pm \sqrt{y + 4}\)

Since \(x \geq 2\), we take the positive root: \(x = \sqrt{y + 4} + 2\)

Thus, \(f^{-1}(x) = \sqrt{x + 4} + 2\).

(c) Given \(a = -\frac{5}{3}\), solve \(f(x) = g(x)\):

\((x - 2)^2 - 4 = -\frac{5}{3}x + 2\)

\(x^2 - 4x + 4 - 4 = -\frac{5}{3}x + 2\)

\(x^2 - 4x + \frac{5}{3}x - 2 = 0\)

\((3x + 2)(x - 3) = 0\)

\(x = 3\) only.

(d) Given \(gg f^{-1}(12) = 62\):

\(f^{-1}(12) = 6\)

\(g(f^{-1}(12)) = 6a + 2\)

\(g(g(f^{-1}(12))) = a(6a + 2) + 2 = 62\)

\(6a^2 + 2a - 60 = 0\)

Solving the quadratic equation gives \(a = \frac{10}{3}\) or \(a = 3\).

(a) To express \(f(x) = x^2 + 2x + 3\) in the form \((x+a)^2 + b\), complete the square:

\(f(x) = (x+1)^2 + 2\).

The range of \(f\) is \(y \geq 2\) since the vertex of the parabola is at \(y = 2\) and it opens upwards.

(b) To find \(f^{-1}(x)\), start with \(y = (x+1)^2 + 2\).

Solving for \(x\), we have:

\(y - 2 = (x+1)^2\)

\(x+1 = \pm \sqrt{y-2}\)

Since \(x \leq -1\), choose the negative root:

\(x = -\sqrt{y-2} - 1\)

Thus, \(f^{-1}(x) = -\sqrt{x-2} - 1\).

(c) Solve \(gf(x) = 13\):

\(g(f(x)) = 2(x^2 + 2x + 3) + 1 = 13\)

\(2x^2 + 4x + 6 + 1 = 13\)

\(2x^2 + 4x - 6 = 0\)

Divide by 2:

\(x^2 + 2x - 3 = 0\)

Factorize:

\((x-1)(x+3) = 0\)

\(x = 1\) or \(x = -3\)

Since \(x \leq -1\), \(x = -3\) only.

(a) To find \(fg(7)\), we first calculate \(g(7)\):

\(g(7) = \frac{4}{7+1} = \frac{4}{8} = \frac{1}{2}\).

Then, calculate \(f(g(7)) = f\left(\frac{1}{2}\right)\):

\(f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right) - 2 = 2 - 2 = 0\).

Thus, \(fg(7) = 0\).

(b) To find \(x\) for which \(f^{-1}(x) = g^{-1}(x)\), we first find the inverses:

\(f(x) = 4x - 2 \Rightarrow y = 4x - 2 \Rightarrow x = \frac{y+2}{4} \Rightarrow f^{-1}(x) = \frac{x+2}{4}\).

\(g(x) = \frac{4}{x+1} \Rightarrow y = \frac{4}{x+1} \Rightarrow y(x+1) = 4 \Rightarrow x = \frac{4}{y} - 1 \Rightarrow g^{-1}(x) = \frac{4}{x} - 1\).

Set \(f^{-1}(x) = g^{-1}(x)\):

\(\frac{x+2}{4} = \frac{4}{x} - 1\).

Multiply through by 4x to clear fractions:

\(x(x+2) = 16 - 4x\).

\(x^2 + 2x = 16 - 4x\).

\(x^2 + 6x - 16 = 0\).

Factor the quadratic:

\((x+8)(x-2) = 0\).

Thus, \(x = -8\) or \(x = 2\).

(a) To find \(fg(x)\), substitute \(g(x) = 2x + 1\) into \(f(x)\):

\(fg(x) = f(g(x)) = (2x+1)^2 + 3\).

(b) Let \(y = (2x+1)^2 + 3\). Solve for \(x\):

\(y - 3 = (2x+1)^2\)

\(2x+1 = \pm \sqrt{y-3}\)

\(x = \frac{1}{2}(\pm \sqrt{y-3} - 1)\)

Since \(x > 0\), choose the positive root:

\((fg)^{-1}(x) = \frac{1}{2}(\sqrt{x-3} - 1)\)

The domain of \((fg)^{-1}\) is \(x > 3\).

(c) Solve \(fg(x) - 3 = gf(x)\):

\((2x+1)^2 + 3 - 3 = 2(x^2+3) + 1\)

\((2x+1)^2 = 2x^2 + 6 + 1\)

\(4x^2 + 4x + 1 = 2x^2 + 7\)

\(2x^2 + 4x - 6 = 0\)

\(2(x+3)(x-1) = 0\)

\(x = 1\)