Answer :
Let's analyze the function [tex]\( f(x) = -2 \log_2 x + 1 \)[/tex] and fill in the blanks step by step.
1. Identify the type of function:
- The given function is [tex]\( f(x) = -2 \log_2 x + 1 \)[/tex].
- This is a logarithmic function transformed by a vertical stretch and a vertical shift.
Therefore, we have:
- The function [tex]\( f(x) \)[/tex] is a logarithmic function.
2. Determine the vertical asymptote:
- For logarithmic functions of the form [tex]\( \log_a x \)[/tex], there is a vertical asymptote at [tex]\( x = 0 \)[/tex].
Thus, we can state:
- The function has a vertical asymptote of [tex]\( x = 0 \)[/tex].
3. Determine the range of the function:
- Logarithmic functions typically have a range of all real numbers, [tex]\( (-\infty, \infty) \)[/tex].
Thus:
- The range of the function is [tex]\( (-\infty, \infty) \)[/tex].
4. Determine the domain of the function:
- Logarithmic functions are only defined for positive values of [tex]\( x \)[/tex], so the domain is [tex]\( (0, \infty) \)[/tex].
Therefore:
- The domain of the function is [tex]\( (0, \infty) \)[/tex].
5. Determine the monotonicity of the function:
- To find out if the function is increasing or decreasing, observe the coefficient of the logarithm:
- The coefficient of the logarithm is [tex]\(-2\)[/tex], which is negative.
- This means the function is decreasing on its domain.
Thus:
- It is decreasing on its domain of [tex]\( (0, \infty) \)[/tex].
6. Analyze the end behavior:
- For the left end behavior [tex]\( (x \to 0^+) \)[/tex]:
- As [tex]\( x \to 0^+ \)[/tex], [tex]\( \log_2 x \to -\infty \)[/tex].
- Given the function [tex]\( -2 \log_2 x + 1 \)[/tex], as [tex]\( \log_2 x \to -\infty \)[/tex], [tex]\( f(x) \)[/tex] will go to [tex]\( +\infty \)[/tex].
Therefore, the end behavior on the LEFT side is as [tex]\( x \to 0^+ \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
- For the right end behavior [tex]\( (x \to \infty) \)[/tex]:
- As [tex]\( x \to \infty \)[/tex], [tex]\( \log_2 x \to \infty \)[/tex].
- Given the function [tex]\( -2 \log_2 x + 1 \)[/tex], as [tex]\( \log_2 x \to \infty \)[/tex], [tex]\( f(x) \)[/tex] will go to [tex]\( -\infty \)[/tex].
Therefore, the end behavior on the RIGHT side is as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
Putting it all together, we have:
The function [tex]\( f(x) \)[/tex] is a logarithmic function with a vertical asymptote of [tex]\( x = 0 \)[/tex]. The range of the function is [tex]\( (-\infty, \infty) \)[/tex], and it is decreasing on its domain of [tex]\( (0, \infty) \)[/tex]. The end behavior on the LEFT side is as [tex]\( x \to 0^+ \)[/tex], [tex]\( f(x) \to \infty \)[/tex], and the end behavior on the RIGHT side is as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
1. Identify the type of function:
- The given function is [tex]\( f(x) = -2 \log_2 x + 1 \)[/tex].
- This is a logarithmic function transformed by a vertical stretch and a vertical shift.
Therefore, we have:
- The function [tex]\( f(x) \)[/tex] is a logarithmic function.
2. Determine the vertical asymptote:
- For logarithmic functions of the form [tex]\( \log_a x \)[/tex], there is a vertical asymptote at [tex]\( x = 0 \)[/tex].
Thus, we can state:
- The function has a vertical asymptote of [tex]\( x = 0 \)[/tex].
3. Determine the range of the function:
- Logarithmic functions typically have a range of all real numbers, [tex]\( (-\infty, \infty) \)[/tex].
Thus:
- The range of the function is [tex]\( (-\infty, \infty) \)[/tex].
4. Determine the domain of the function:
- Logarithmic functions are only defined for positive values of [tex]\( x \)[/tex], so the domain is [tex]\( (0, \infty) \)[/tex].
Therefore:
- The domain of the function is [tex]\( (0, \infty) \)[/tex].
5. Determine the monotonicity of the function:
- To find out if the function is increasing or decreasing, observe the coefficient of the logarithm:
- The coefficient of the logarithm is [tex]\(-2\)[/tex], which is negative.
- This means the function is decreasing on its domain.
Thus:
- It is decreasing on its domain of [tex]\( (0, \infty) \)[/tex].
6. Analyze the end behavior:
- For the left end behavior [tex]\( (x \to 0^+) \)[/tex]:
- As [tex]\( x \to 0^+ \)[/tex], [tex]\( \log_2 x \to -\infty \)[/tex].
- Given the function [tex]\( -2 \log_2 x + 1 \)[/tex], as [tex]\( \log_2 x \to -\infty \)[/tex], [tex]\( f(x) \)[/tex] will go to [tex]\( +\infty \)[/tex].
Therefore, the end behavior on the LEFT side is as [tex]\( x \to 0^+ \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
- For the right end behavior [tex]\( (x \to \infty) \)[/tex]:
- As [tex]\( x \to \infty \)[/tex], [tex]\( \log_2 x \to \infty \)[/tex].
- Given the function [tex]\( -2 \log_2 x + 1 \)[/tex], as [tex]\( \log_2 x \to \infty \)[/tex], [tex]\( f(x) \)[/tex] will go to [tex]\( -\infty \)[/tex].
Therefore, the end behavior on the RIGHT side is as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
Putting it all together, we have:
The function [tex]\( f(x) \)[/tex] is a logarithmic function with a vertical asymptote of [tex]\( x = 0 \)[/tex]. The range of the function is [tex]\( (-\infty, \infty) \)[/tex], and it is decreasing on its domain of [tex]\( (0, \infty) \)[/tex]. The end behavior on the LEFT side is as [tex]\( x \to 0^+ \)[/tex], [tex]\( f(x) \to \infty \)[/tex], and the end behavior on the RIGHT side is as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
