a) The vertical asymptote occurs when the denominator of the fraction becomes zero. In the given function:
\[ f(x) = 3 + \frac{15}{3x - 3} \]
The denominator is \( 3x - 3 \), which equals 0 when:
\[ 3x - 3 = 0 \implies x = 1 \]
Therefore, the equation of the vertical asymptote is:
\[ x = 1 \]
b) To find \( f^{-1}(13) \), we solve for \( x \) in the equation \( f(x) = 13 \):
\[ 13 = 3 + \frac{15}{3x - 3} \]
\[ 10(3x - 3) = 15 \]
\[ 30x = 45 \]
\[ x = \frac{45}{30} = \frac{3}{2} \]
Thus, \( f^{-1}(13) = \frac{3}{2} \).
Closea) To find \( f^{-1}(3) \), we solve for \( x \) in the equation \( f(x) = 3 \):
\[ f(x) = \frac{3}{\sqrt{2x+1}} - 2 \]
Set \( f(x) = 3 \):
\[ 3 = \frac{3}{\sqrt{2x+1}} - 2 \]
\[ x = -\frac{16}{50} = -\frac{8}{25} \]
Thus, \( f^{-1}(3) = -\frac{8}{25} \).
b) The range of \( f(x) \) is determined by the possible values of \( \frac{3}{\sqrt{2x+1}} - 2 \) for \( x > -\frac{1}{2} \):
1. As \( x \to -\frac{1}{2}^+ \), \( \sqrt{2x+1} \to 0^+ \), making \( \frac{3}{\sqrt{2x+1}} \to \infty \). Thus, \( f(x) \to \infty - 2 = \infty \).
2. As \( x \to \infty \), \( \sqrt{2x+1} \to \infty \), making \( \frac{3}{\sqrt{2x+1}} \to 0^+ \). Thus, \( f(x) \to 0 - 2 = -2 \).
Therefore, the range of \( f(x) \) is: \( (-2, \infty) \)
c) The domain of \( f^{-1}(x) \) is the range of \( f(x) \), which is: \( (-2, \infty) \)
Closea)
b) \( (-0.721,4.24) \)
c) The vertical asymptote is when the denominator is zero, so \( x = 0 \).
d) The best way to solve it is to draw a straight line at \( y = 5 \) and find all intersecting points:
So the x-coordinates of these points are: \( -1, \ -0.5, \ 1.5 \).
Closea) To find the range of \( l(x) \), evaluate the function \( l(x) = \frac{300}{x^2+1} - 2 \) for the domain \( 3 \leq x \leq 20 \):
1. As \( x \to 3 \), \( x^2 + 1 = 10 \), and \( \frac{300}{10} = 30 \). Therefore, \( l(3) = 30 - 2 = 28 \).
2. As \( x \to 20 \), \( x^2 + 1 = 401 \), and \( \frac{300}{401} \approx 0.748 \). Therefore, \( l(20) \approx 0.748 - 2 = -1.252 \).
Thus, the range of \( l(x) \) is: \( (-1.25, 28) \)
b) To find \( l^{-1}(4) \), solve \( l(x) = 4 \):
\[ \frac{300}{x^2+1} - 2 = 4 \]
\[ \frac{300}{x^2+1} = 6 \]
\[ x^2+1 = \frac{300}{6} = 50 \]
\[ x^2 = 49 \implies x = \pm 7 \]
Since \( x \geq 3 \), \( x = 7 \). Thus, \( l^{-1}(4) = 7 \).
c) The interpretation of \( l^{-1}(4) = 7 \) in this context is that when the length of the road is 4 kilometers, the mass of concrete used is 7 kilograms.
d) The range of \( l^{-1}(x) \) is the domain of \( l(x) \), which is: \( [3, 20] \)
Closea)
\[ f(2) = 1 + 2^2 = 5\]
\[ g(2) = \frac{1}{1+2} = \frac{1}{3}\]
b) The domain of \( f \) is \( \mathbb{R} \), but \( g \) is not defined for \( x = -1 \) (because then we would divide by zero), so we have to exclude it from the domain of \( f + g \).
c) The range of \( f(x) \) is \( y > 1 \), but \( g(x) \) reaches values \( \leq 1 \), and the range of \( g(x) \) is \( y \neq 0 \), but \( f(x) \) reaches 0, so the total range is \( \mathbb{R} \).
We can also see this graphically:
d)
\[ (f+g)(2) = f(2) + g(2) = 5 + \frac{1}{3} = \frac{16}{3}\]
Closea) 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:
Close
a) We use one of the easily read coordinates on the graph, for example (0,2):
\[ f(2) = \frac{(2+k)^2}{2}=0 \implies (2+k)^2 = 0 \implies k = -2\]
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) To answer this question we use the fact that the domain of \( f(x) \) is equal to the range of its inverse, and vice versa. Therefore, the first point we're looking for, \( f^{-1}(2) \), is at the point where \( y = 2 \) for the original function. By looking at the graph, we can see that \( y = 2 \) occurs at two points, \( x = 0 \) and \( x = 4 \).
Similarly, \( f^{-1}(0) \) will happen when \( y = 0 \) for the original function. We can clearly see that for the original function, \( y = 0 \) at \( x = 2 \).
c) We can see that the graph is symmetrical with respect to \( x = 2 \). Alternatively, we can rewrite \( f(x) \) as \( f(x) = \frac{1}{2}x^2 - 2x + 2 \) , which is a quadratic function with axis of symmetry at:
\[ x = - \frac{b}{2a} = \frac{2}{2 \cdot \frac{1}{2}} = 2 \]
Closea) It can be clearly seen from the function's graph that the vertical asymptote is at \( x = 1 \), and the horizontal asymptote is at \( y = 2 \).
b) Since we don't know the equation of the function, we can't directly plug points into the inverse formula. However, knowing the relationship between the domain and range of inverse functions, we can find those points by looking at the graph. Therefore, taking the first point as an example, \( f^{-1}(0) \) is actually the point for which \( y = 0 \) for the function \( f(x) \). As we can see, the x-coordinate for this point is \( x = 2 \).
For the second point, \( f^{-1}(2) \) means we are looking for the point when \( y = 2 \) in \( f(x) \). As we can see, \( y = 2 \) is undefined for the original function since this is the value of one of the asymptotes.
The third point, \( f^{-1}(-2) \), means we are looking for when \( y = -2 \) for the original function. At \( y = -2 \), we can see that \( x = 0.5 \), which means that \( f^{-1}(-2) = 0.5 \).
Lastly, using the same approach, when \( y = 4 \) for the original function, then \( x = 2 \), meaning that \( f^{-1}(4) = 2 \).
Closea)
\[ f(2) = 0.5 * 2 + 2 = 3\]
\[ f(4) = 0.5 * 4 + 2 = 4\]
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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