The inverse function \(\displaystyle f^{-1}(x)\) of a function \(\displaystyle f(x)\) can be found as follows:

Replace \(\displaystyle f(x)\) with \(\displaystyle y\):

\(\displaystyle f(x) = \sqrt{4x- 9}\)

\(\displaystyle y= \sqrt{4x- 9}\)

Switch the positions of \(\displaystyle y\) and \(\displaystyle x\):

\(\displaystyle x= \sqrt{4y- 9}\)

or

\(\displaystyle \sqrt{4y- 9} = x\)

Solve for \(\displaystyle y\). This can be done as follows:

Square both sides:

\(\displaystyle \left (\sqrt{4y- 9} \right ) ^{2}= x ^{2}\)

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

Add 9 to both sides:

\(\displaystyle 4y- 9+9 = x ^{2}+9\)

\(\displaystyle 4y = x ^{2}+9\)

Multiply both sides by \(\displaystyle \frac{1}{4}\), distributing on the right:

\(\displaystyle \frac{1}{4} \cdot 4y = \frac{1}{4} \cdot \left (x ^{2}+9 \right )\)

\(\displaystyle y = \frac{1}{4} \cdot x ^{2}+ \frac{1}{4} \cdot 9\)

\(\displaystyle y = \frac{1}{4} x ^{2}+ \frac{9}{4}\)

Replace \(\displaystyle y\) with \(\displaystyle f^{-1}(x)\):

\(\displaystyle f^{-1}(x)= \frac{1}{4} x ^{2}+ \frac{9}{4}\)

The statement is false. Look for the point on the graph of \(\displaystyle f(x)\) with \(\displaystyle x\)-coordinate \(\displaystyle \frac{1}{2}\) by going right \(\displaystyle \frac{1}{2}\) unit, then moving up and noting the \(\displaystyle y\)-value, as follows:

\(\displaystyle f \left ( \frac{1}{2} \right ) = \frac{3}{2}\), so the statement is false.

The \(\displaystyle y\)-intercept of the graph of a function is the point at which it intersects the \(\displaystyle y\)-axis (the vertical axis). That point is marked on the diagram below:

The point is about one and three-fourths units above the origin, making the coordinates of the \(\displaystyle y\)-intercept \(\displaystyle \left ( 0, 1\frac{3}{4}\right )\).

This is the composition of two functions. By definition, \(\displaystyle (g \circ f )(x) = g \left [ f(x)\right ]\).  To find the range of \(\displaystyle (g \circ f )(x)\), we need to find the values of this function for each value in the domain of \(\displaystyle f(x)\). Since \(\displaystyle (g \circ f )(x) = g \left [ f(x)\right ]\), this is equivalent to evaluating \(\displaystyle g(x)\) for each value in the range of \(\displaystyle f(x)\), as follows:

\(\displaystyle g(x) = 3x- 5\)

Range value: 3 

\(\displaystyle g(3) = 3 (3)- 5 = 9 - 5 = 4\)

Range value: 5

\(\displaystyle g(5) = 3 (5)- 5 = 15 - 5 = 10\)

Range value: 8

\(\displaystyle g(8) = 3 (8)- 5 = 24 - 5 = 19\)

Range value: 13

\(\displaystyle g(13) = 3 (13)- 5 = 39 - 5 = 34\)

Range value: 21

\(\displaystyle g(21) = 3 (21)- 5 = 63 - 5 = 58\)

The range of \(\displaystyle g(x)\) on the set of range values of \(\displaystyle f(x)\) - and consequently, the range of \(\displaystyle \left (g \circ f \right ) (x)\) - is the set \(\displaystyle \left \{4 , 10, 19, 34, 58\right \}\). Of  the five choices, only 45 does not appear in this set; this is the correct choice.

Evaluate the expression \(\displaystyle (-1)^{k}(3k+1)\) for \(\displaystyle k = 2, 3, 4, 5\), then add the four numbers:

\(\displaystyle a_{k} = (-1)^{k}(3k+1)\)

\(\displaystyle a_{2} = (-1)^{2}(3 \cdot 2 +1) = 1 (6+ 1) = 7\)

\(\displaystyle a_{3} = (-1)^{3}(3 \cdot 3 +1) = -1 (9+ 1) = -10\)

\(\displaystyle a_{4} = (-1)^{4}(3 \cdot 4 +1) = 1 (12+ 1) = 13\)

\(\displaystyle a_{5} = (-1)^{5}(3 \cdot 5 +1) =- 1 (15+ 1) = -16\)

\(\displaystyle \sum_{k =2}^{5} (-1)^{k}(3k+1) = 7 +( -10) + 13 + (-16 ) = -6\) 

Evaluate the expression \(\displaystyle \sum_{k =3}^{7} (-2)^{k}\) for \(\displaystyle k = 3, 4, 5, 6, 7\), then add the five numbers:

\(\displaystyle a_{n} = (-2)^{k}\)

\(\displaystyle a_{3} = (-2)^{3} = -8\)

\(\displaystyle a_{4} = (-2)^{4 } =16\)

\(\displaystyle a_{5} = (-2)^{5 } =-32\)

\(\displaystyle a_{6} = (-2)^{6 } =64\)

\(\displaystyle a_{7} = (-2)^{7 } =-128\)

\(\displaystyle \sum_{k =3}^{7} (-2)^{k} = -8 + 16 + (-32)+ 64 +( -128 ) = -88\)

\(\displaystyle (f \circ g ) \left ( \frac{11}{4} \right ) =f\left ( g \left ( \frac{11}{4} \right ) \right )\), so, first, evaluate \(\displaystyle g \left ( \frac{11}{4} \right )\) by substitution:

\(\displaystyle g(x) = \left \lceil \frac{2}{3}x- \frac{1}{2} \right \rceil\)

\(\displaystyle g \left ( \frac{11}{4} \right ) = \left \lceil \frac{2}{3} \left ( \frac{11}{4} \right ) - \frac{1}{2} \right \rceil\)

\(\displaystyle = \left \lceil \frac{11}{6} - \frac{1}{2} \right \rceil\)

\(\displaystyle = \left \lceil \frac{4}{3} \right \rceil\)

\(\displaystyle = 2\)

\(\displaystyle (f \circ g ) \left ( \frac{11}{4} \right ) =f\left ( g \left ( \frac{11}{4} \right ) \right ) = f(2)\), so evaluate \(\displaystyle f(2)\) by substitution.

\(\displaystyle f(x) = \left \lfloor \frac{1}{2}x- \frac{2}{3} \right \rfloor\)

\(\displaystyle f(2) = \left \lfloor \frac{1}{2} (2)- \frac{2}{3} \right \rfloor\)

\(\displaystyle = \left \lfloor1 - \frac{2}{3} \right \rfloor\)

\(\displaystyle = \left \lfloor \frac{1}{3} \right \rfloor\)

\(\displaystyle = 0\),

the correct response.

\(\displaystyle (f \circ g )(3.4 ) = f(g(3.4))\), so first, evaluate \(\displaystyle g(3.4 )\) using substitution:

\(\displaystyle g (x) = \left \lfloor -2.1 x + 9.3 \right \rfloor\)

\(\displaystyle g (3.4) = \left \lfloor -2.1 (3.4) + 9.3 \right \rfloor\)

\(\displaystyle = \left \lfloor -7.14+ 9.3 \right \rfloor\)

\(\displaystyle = \left \lfloor 2.16 \right \rfloor\)

\(\displaystyle = 2\)

\(\displaystyle (f \circ g )(3.4 ) = f(g(3.4)) = f(2)\), so evaluate \(\displaystyle f(2)\) using substitution:

\(\displaystyle f(x) = \left \lceil 2.2x +6.3 \right \rceil\)

\(\displaystyle = \left \lceil 2.2 (2) +6.3 \right \rceil\)

\(\displaystyle = \left \lceil 4.4 +6.3 \right \rceil\)

\(\displaystyle = \left \lceil 10.7 \right \rceil\)

\(\displaystyle = 11\),

the correct response.

By the Factor Theorem, \(\displaystyle x- c\) is a zero of a polynomial \(\displaystyle P(x)\) if and only if \(\displaystyle P(c) = 0\). Here, \(\displaystyle c = 2\), so evaluate the polynomial, in terms of \(\displaystyle k\), for \(\displaystyle x=2\) by substituting 2 for \(\displaystyle x\):

\(\displaystyle 2 \cdot 2^{5} + 3 \cdot 2^{2}- 4 \cdot 2 + k\)

\(\displaystyle = 2 \cdot 32 + 3 \cdot 4 - 4 \cdot 2 + k\)

\(\displaystyle =64 + 12 - 8 + k\)

\(\displaystyle =68 + k\)

Set this equal to 0:

\(\displaystyle 68 + k = 0\)

\(\displaystyle k = -68\)