By definition, \(\displaystyle (f-g)(19) = f(19)- g(19)\); simply evaluate \(\displaystyle f(19)\) and \(\displaystyle g(19)\) by substituting 19 for \(\displaystyle x\) in both definitions, and subtract:

\(\displaystyle f(x) = 4x- 17\)

\(\displaystyle f(19) = 4 \cdot 19 - 17 = 76 - 17 = 59\)

\(\displaystyle g(x) = \frac{1}{2}x+ 13\)

\(\displaystyle g(19) = \frac{1}{2} \cdot 19 + 13 = 9\frac{1}{2} + 13 = 22 \frac{1}{2}\)

\(\displaystyle (f-g)(19) = f(19)- g(19) = 59 - 22 \frac{1}{2} = 36 \frac{1}{2}\)

By definition, \(\displaystyle (f \circ g ) (x) = f [g(x)]\).

Replacing \(\displaystyle g(x)\) with its definition, we get

\(\displaystyle (f \circ g ) (x) = f \left ( \frac{1}{2}x+ 13 \right )\)

In the definition of \(\displaystyle f ( x)\), replace \(\displaystyle x\) with \(\displaystyle \frac{1}{2}x+ 13\) and simplify the expression:

\(\displaystyle f(x) = 4x- 17\)

\(\displaystyle f \left ( \frac{1}{2}x+ 13 \right )= 4 \left ( \frac{1}{2}x+ 13 \right )- 17\)

\(\displaystyle = 2 x+ 52 - 17\)

\(\displaystyle = 2 x+ 35\)

Therefore, 

\(\displaystyle (f \circ g ) (x) = 2 x+ 35\)

If 

\(\displaystyle (f \circ g) (c) = 10\),

then 

\(\displaystyle 2 c+35 = 10\)

Solve for \(\displaystyle c\):

\(\displaystyle 2 c+35-35 = 10 - 35\)

\(\displaystyle 2 c = -25\)

\(\displaystyle \frac{1}{2} \cdot 2 c = \frac{1}{2} \cdot (-25)\)

\(\displaystyle c = -12 \frac{1}{2}\)

By definition, \(\displaystyle (g \circ f ) (x) = g [f(x)]\).

Replacing \(\displaystyle f(x)\) with its definition, we get

\(\displaystyle (g \circ f ) (x) = g ( 4x- 17)\)

In the definition of \(\displaystyle g ( x)\), replace \(\displaystyle x\) with \(\displaystyle 4x-17\) and simplify the expression:

\(\displaystyle g(x) = \frac{1}{2}x+ 13\)

\(\displaystyle g(4x-17 ) = \frac{1}{2} (4x-17 )+ 13\)

\(\displaystyle = 2 x-8 \frac{1}{2} + 13\)

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

Therefore, 

\(\displaystyle (g \circ f ) (x) = 2 x+4 \frac{1}{2}\)

If 

\(\displaystyle (g \circ f ) (c) = 10\),

then 

\(\displaystyle 2 c+4 \frac{1}{2} = 10\)

Solve for \(\displaystyle c\):

\(\displaystyle 2 c+4 \frac{1}{2} - 4 \frac{1}{2} = 10 - 4 \frac{1}{2}\)

\(\displaystyle 2 c = 5 \frac{1}{2}\)

\(\displaystyle \frac{1}{2} \cdot 2 c = \frac{1}{2} \cdot 5 \frac{1}{2}\)

\(\displaystyle c = 2 \frac{3}{4}\)

To obtain the definition of the function \(\displaystyle (f-g)(x)\), subtract the expressions that define the individual functions \(\displaystyle f(x)\) and \(\displaystyle g(x)\):

\(\displaystyle f(x) = 4x- 17\)

\(\displaystyle g(x) = \frac{1}{2}x+ 13\)

\(\displaystyle (f+g)(x)= (4x- 17 )- \left ( \frac{1}{2}x+ 13 \right )\)

\(\displaystyle = 4x- \frac{1}{2}x- 17 -13\)

\(\displaystyle = \frac{7}{2}x- 30\)

\(\displaystyle (f+g) (c) = 19\), so

\(\displaystyle \frac{7}{2}c- 30 = 19\)

Solve for \(\displaystyle c\); add 30:

\(\displaystyle \frac{7}{2}c- 30 + 30 = 19 + 30\)

\(\displaystyle \frac{7}{2}c = 49\)

Multiply by \(\displaystyle \frac{2}{7}\):

\(\displaystyle \frac{2} {7}\cdot \frac{7}{2}c = \frac{2} {7}\cdot 49\)

\(\displaystyle c =14\)

To obtain the definition of the function \(\displaystyle (f+g)(x)\), add the expressions that define the individual functions \(\displaystyle f(x)\) and \(\displaystyle g(x)\)"

\(\displaystyle f(x) = 4x- 17\)

\(\displaystyle g(x) = \frac{1}{2}x+ 13\)

\(\displaystyle (f+g)(x)= (4x- 17 )+ \left ( \frac{1}{2}x+ 13 \right )\)

\(\displaystyle = 4x+ \frac{1}{2}x- 17 + 13\)

\(\displaystyle = \frac{9}{2}x- 4\)

\(\displaystyle (f+g) (c) = 19\), so

\(\displaystyle \frac{9}{2}c- 4 = 19\);

Solve for \(\displaystyle c\); add 4;

\(\displaystyle \frac{9}{2}c- 4+ 4 = 19 + 4\)

\(\displaystyle \frac{9}{2}c = 23\)

Multiply by \(\displaystyle \frac{2}{9}\):

\(\displaystyle \frac{2}{9} \cdot \frac{9}{2}c =\frac{2}{9} \cdot 23\)

\(\displaystyle c =\frac{46}{9} = 5 \frac{1}{9}\)

A function of the form \(\displaystyle f(x) = m x + b\) is a linear function and is either constantly increasing or constantly decreasing. Therefore, \(\displaystyle f(x) = 7x - 17\) has its minimum and maximum values at the endpoints of its domain.

We evaluate \(\displaystyle f(-4)\) and \(\displaystyle f(8)\) by substitution, as follows:

\(\displaystyle f(x) = 7x - 17\)

\(\displaystyle f(-4) = 7 (-4) - 17 = -28 - 17 = -45\)

\(\displaystyle f(8) = 7 (8) - 17 = 56 - 17 = 39\)

The range of the function on the domain to which it is restricted is \(\displaystyle \left [ -45, 39 \right ]\).

Since the piecewise-defined function \(\displaystyle f(x)\) is defined two different ways, one for negative numbers and one for nonnegative numbers, examine both definitions and determine each partial range separately;  the union of the partial ranges will be the overall range.

If \(\displaystyle x < 0\), then 

\(\displaystyle f(x ) = 2x- 7\) 

Since 

\(\displaystyle x < 0\),

applying the properties of inequality,

\(\displaystyle 2 x < 2 \cdot 0\)

\(\displaystyle 2 x < 0\)

\(\displaystyle 2 x - 7 < 0 - 7\)

\(\displaystyle 2 x - 7 < - 7\)

\(\displaystyle f(x) < -7\)

Therefore, on the portion of the domain comprising nonpositive numbers, the partial range of \(\displaystyle f(x)\) is the set \(\displaystyle (-\infty , -7)\).

If \(\displaystyle x \ge 0\), then 

\(\displaystyle f(x ) = 4x-14\) 

Since 

\(\displaystyle x \ge 0\),

applying the properties of inequality,

\(\displaystyle 4 x \ge 4 \cdot 0\)

\(\displaystyle 4 x \ge 0\)

\(\displaystyle 4 x -14 \ge 0 - 14\)

\(\displaystyle 4 x -14 \ge - 14\)

\(\displaystyle f(x) \ge - 14\)

Therefore, on the portion of the domain comprising positive numbers, the partial range of \(\displaystyle f(x)\) is the set \(\displaystyle [-14, \infty)\).

The overall range is the union of these partial ranges, which is \(\displaystyle (-\infty , \infty)\).

First, we must determine whether \(\displaystyle f^{-1} (x)\) exists.

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place. 

\(\displaystyle f^{-1} (x)\) exists if and only if, if \(\displaystyle f(a) = f(b)\), then \(\displaystyle a = b\)- or, equivalently, if there does not exist \(\displaystyle a\) and \(\displaystyle b\) such that \(\displaystyle a \ne b\), but \(\displaystyle f(a) = f(b)\). This will happen on any interval on which the graph of \(\displaystyle f\) constantly increases or constantly decreases, but if the graph changes direction on an interval, there will be \(\displaystyle a \ne b\) such that \(\displaystyle f(a) = f(b)\) on this interval. The key is therefore to determine whether the interval to which the domain is restricted contains the vertex.

The \(\displaystyle x\)-coordinate of the vertex of the parabola of the function

\(\displaystyle f(x) = ax^{2}+ bx+c\)

is \(\displaystyle x = - \frac{b}{2a}\).

The \(\displaystyle x\)-coordinate of the vertex of the parabola of \(\displaystyle f(x) = x^{2}- 8x - 4\) can be found by setting \(\displaystyle a = 1, b = -8\):

\(\displaystyle x = - \frac{-8}{2 \cdot 1} = \frac{8}{2} =4\).

The vertex of the graph of \(\displaystyle f\) without its domain restriction is at the point with \(\displaystyle x\)-coordinate 4. Since \(\displaystyle 4 \in \left [ 0, \infty \right )\), the vertex is in the interior of the domain; as a consequence, \(\displaystyle f^{-1} (x)\) does not exist on \(\displaystyle \left [ 0, \infty \right )\).

A function of the form \(\displaystyle f(x) = m x + b\) is a linear function and is either constantly increasing or constantly decreasing. Therefore, we can simply note that if

\(\displaystyle x \ge 5\),

as stated in the domain, then, multiplying both sides by \(\displaystyle -4\), remembering to switch the symbol since we are multiplying by a negative number:

\(\displaystyle -4 \cdot x \le -4 \cdot 5\)

\(\displaystyle -4 x \le -20\)

Add 12 to both sides:

\(\displaystyle -4 x + 12 \le -20 + 12\)

\(\displaystyle -4 x + 12 \le -8\)

Replacing, we see that 

\(\displaystyle f(x) \le -8\),

so the range of \(\displaystyle f(x)\) is \(\displaystyle \left ( -\infty, 8 \right ]\).

If \(\displaystyle x \ge 1\), it follows by applying the properties of inequality that:

\(\displaystyle 2 \cdot x \ge 2 \cdot 1\)

\(\displaystyle 2x \ge 2\)

\(\displaystyle 2x + 3 \ge 2 + 3\)

\(\displaystyle 2x + 3 \ge 5\)

Multiply both sides by \(\displaystyle \frac{1}{5(2x+3)}\), which must be positive by closure:

\(\displaystyle (2x + 3 )\cdot \frac{1}{5(2x+3)} \ge 5 \cdot \frac{1}{5(2x+3)}\)

\(\displaystyle \frac{1}{5 } \ge \frac{1}{2x+3}\), 

that is, 

\(\displaystyle \frac{1}{2x+3} \le \frac{1}{5 }\)

Also,by closure, \(\displaystyle \frac{1}{2x+3} > 0\), so

\(\displaystyle 0 < \frac{1}{2x+3} \le \frac{1}{5 }\)

\(\displaystyle 0 < f(x) \le \frac{1}{5 }\)

This makes the correct range \(\displaystyle \left (0, \frac{1}{5} \right ]\).