Detailed Answer

a)

\[ f(2) = 0.5 * 2 + 2 = 3\]

\[ f(4) = 0.5 * 4 + 2 = 4\]

b) This can be easily done in GDC resulting in the following graph.

c) This question can either be answered by looking at the original graph of f(x) or by calculating the inverse function:

\[ y = \frac{1}{2}x + 2\]

\[ x = \frac{1}{2}y + 2\]

\[ x - 2 = \frac{1}{2}y\]

\[ y = 2x - 4\]

Then, by plugging those two points into this inverse function we get that:

\[ f^{-1}(2) = 2*2 - 4 = 0\]

\[ f^{-1}(3.5) = 2*3.5 - 4 = 3\]

d) This question can be solved in multiple ways.

Firstly, from the previous question, we know the coordinates of two points of this line, and we can draw it.

Alternatively, we can draw the line \( y=x \), and draw the symmetry of f with regards to it.

We can also place the point with coordinates (3,2) and draw the line going through it and (4,4).

Or you can just put it into GDC 😉

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Detailed Answer

a) The domain is defined if: \(\sqrt{4-x} \neq 0\) .

\(\sqrt{4-x} \neq 0\) is defined for \( x \leq 4 \), only reaching 0 at \( x = 4 \). So the domain of \( f \) is \( \{ x \in \mathbb{R} \mid x < 4 \} \).

As for the range, \(\frac{4}{\sqrt{4-x}}\) is always greater than 0, and we have to subtract 2 (as the function subtracts 2 from the fraction). Therefore, the range is for \( y > -2 \).

b) The asymptotes can be clearly seen by looking at the graph of the function or analyzing the domain and range from part (a). Therefore, the vertical asymptote is at \( x = 4 \), and the horizontal asymptote is at \( y = -2 \).

c) The inverse can be easiest found by first flipping the order of \( y \) and \( x \) in the original function, and then solving for \( y \):

\[ y = \frac{4}{\sqrt{4-x}} - 2 \]

\[ x = \frac{4}{\sqrt{4-y}} - 2 \]

\[ x + 2 = \frac{4}{\sqrt{4-y}} \]

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

\[ \sqrt{4-y} = \frac{4}{x+2} \]

\[ 4-y = \left(\frac{4}{x+2}\right)^2 \]

\[ y = 4 - \left(\frac{4}{x+2}\right)^2 \]

d) The domain and range of the inverse can be easily found by applying the rule stating that the domain of \( f(x) \) is equal to the range of \( f^{-1}(x) \) and the range of \( f(x) \) is equal to the domain of \( f^{-1}(x) \). By applying this rule, we can easily find that the domain of the inverse is \( x > -2 \), and the range is \( y < 4 \).

e) and f) Both of these subquestions can be answered using the graph below:

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Detailed Answer

a)

(a)	Algebraically	Graphically
Domain	We need the number under the square root to be ≥ 0, so 4𝑥 − 8 ≥ 0. This means the domain is {𝑥|𝑥 ∈ R, 𝑥 ≥ 2}.	The smallest value of x with an image is 2,so the domain is 𝑥 ≥ 2.
Range	\(\sqrt{4x-8}\) is always ≥ 0, so \(\sqrt{4x-8} - 2\) is always ≥ −2. The range is {𝑦|𝑦 ∈ R, 𝑦 ≥ 2}.	The values for y start at -2, so the range is 𝑦 ≥ −2.

b)

(b)	Algebraically	Graphically
Domain	We can’t divide by zero, so we need 𝑥 ≠ 3. The domain is {𝑥|𝑥 ∈ R,𝑥 ≠ 3}.	There is an asymptote for x=3, where the function has no value. So we know the domain is all of R where 𝑥 ≠ 3.
Range	The only number not in the range of \(\frac{1}{3-x}\)is 0. Since we add 2 to all values, the range of the function is all of R except x = 2. The range is: {𝑦|𝑦 ∈ R, 𝑦 ≠ 2}.	All values of y are reached except y=2 where there is another asymptote. So the range is R where 𝑦 ≠ 2.

c)

(c)	Algebraically	Graphically
Domain	We can’t divide by zero, so we must have 𝑥3 − 1 ≠ 0, which means 𝑥 ≠ 1. The domain is {𝑥|𝑥 ∈ R, 𝑥 ≠ 1}.	It’s not clear here if there is an asymptote at x = 1 or If the graph has a “peak” and goes back down, which is why it’s important to also do the algebraic approach. From the graph, the domain can be R or R with 𝑥 ≠ 1.
Range	The denominator is always positive, so \(\frac{3}{{(x^3-1)}^2}\) hits all positive values except 0. This means the range is {𝑦|𝑦 ∈ R, 𝑦 > 0}.	All values strictly above 0 appear to be reached, so the range is 𝑦 > 0.

d)

(d)	Algebraically	Graphically
Domain	We can’t divide by zero, so we must have 𝑥2 − 1 ≠ 0, which means 𝑥 ≠ 1and 𝑥 ≠ −1.The domain is {𝑥|𝑥 ∈ R, 𝑥 ∉ {−1,1}}.	There are two asymptotes for x = -1 and x = 1, so the domain is all of R where 𝑥 ≠ −1 and 𝑥 ≠ 1.
Range	Like before, denominator always positive and fraction always non null, so the range is {𝑦|𝑦 ∈ R, 𝑦 > 0}.	All values strictly above 0 appear to be reached again, so again the range is 𝑦 > 0.

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Detailed Answer

a) \( f(1) = 0 \)

b) \( f(2) = 3 \)

c) To find it we have to look for the x-value when \( y = 3 \), so 2.

d) To find \( (f \circ f)(1) \), we first find \( f(1) \), because we know that \( (f \circ f)(1) = f(f(1)) \):

\[ f(1) = 0 \]

Next, we find \( f(f(1)) \):

\[ f(f(1)) = f(0) \]

\[ f(0) -1 \]

Therefore, \( (f \circ f)(1) = f(f(1)) = f(0) = -1 \).

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Detailed Answer

a) To find the gradient of the line \( [AB] \), we use the formula for the gradient:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Substitute the given points \( A(3, 12) \) and \( B(-2, -3) \):

\[ m = \frac{-3 - 12}{-2 - 3} = \frac{-15}{-5} = 3 \]

So, the gradient of the line \( [AB] \) is \( 3 \).

b)

\[ y = mx + c \]

We have the gradient \( m = 3 \). Using the point \( A(3, 12) \) we can find \( c \):

\[ 12 = 3 * 3 + c \]

\[ 12 = 9 + c \]

\[ c = 3 \]

Thus, the equation of the line \( [AB] \) is:

\[ y = 3x + 3 \]

c) Line \( L_2 \) is perpendicular to \( [AB] \). The gradient of a line perpendicular to another is the negative reciprocal of the original line's gradient. Given the gradient of \( [AB] \) is \( 3 \), the gradient of \( L_2 \) is:

\[ m_{L_2} = -\frac{1}{3} \]

d) Points \( C(9, 1) \) and \( D(x, 5) \) lie on \( L_2 \). The gradient of \( L_2 \) is \( -\frac{1}{3} \). Using the gradient formula between points \( C(9, 1) \) and \( D(x, 5) \):

\[ \frac{5 - 1}{x - 9} = -\frac{1}{3} \]

Solving for \( x \):

\[ \frac{4}{x - 9} = -\frac{1}{3} \]

\[ 4 = -\frac{x - 9}{3} \]

\[ 12 = -(x - 9) \]

\[ 12 = -x + 9 \]

\[ x = -3 \]

Thus, the value of \( x \) is \( -3 \).

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Detailed Answer

a) We find the intercepts immediately by reading the equations. We can also replace \(x\) by \(0\) to be sure.

b) By equalizing the functions we get that the point of interesection is at \((1,2)\).

c) If \(L_{3}\) is perpendicular to the y-axis, then it is a horizontal line with an equation of the form \(y = c\). It goes through \(M\), so \(y = 2\), as \(y_{M} = 2\).

d) \(L_{3}\) is always equal to \(2\). It is parallel to the x-axis and never intersects it.

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Detailed Answer

a) To find \( g^{-1}(x) \), let's start with the equation for \( g(x) \):

\[ y = \frac{x + 3}{5} \]

Swap \( x \) and \( y \) to solve for \( y \):

\[ x = \frac{y + 3}{5} \]

\[ y = 5x - 3 \]

b) To find \( (f \circ g)(x) \), we find \( f(g(x)) \).

\[ g(x) = \frac{x + 3}{5} \]

\[ f(g(x)) = f\left(\frac{x + 3}{5}\right) = 25 \left( \frac{x + 3}{5} \right) + 10 \]

\[ (f \circ g)(x) = 5x + 25 \]

c) From part (a), we have:

\[ g^{-1}(x) = 5x - 3 \]

Substitute \( g^{-1}(x) \) into \( f(x) \):

\[ f(5x - 3) = 25(5x - 3) + 10 \]

\[ = 125x - 75 + 10 = 125x - 65 \]

d)

\[ 125q - 65 = 185 \]

\[ 125q = 250 \]

\[ q = 2 \]

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Detailed Answer

a)

\[ (h \circ g)(x) = 3(2e^{x+1})^3 + 2 = 3(8e^{3x+3}) + 2 = 24e^{3(x+1)} + 2\]

b) By inputting the two functions into the GDC we find the point of interesection at (-2.33, 2.44):

c) Again, in the GDC we can find that the minimum is at (0, -3):

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