IDNLearn.com is your trusted platform for finding reliable answers. Our platform provides detailed and accurate responses from experts, helping you navigate any topic with confidence.
Sagot :
Let's analyze the function [tex]\( g \)[/tex] step by step:
### Step 1: Define the function
The function [tex]\( g \)[/tex] is defined piecewise as:
[tex]\[ g(x) = \begin{cases} \left(\frac{3}{4}\right)^x & \text{for } x < 0 \\ -x^2 & \text{for } x \geq 0 \end{cases} \][/tex]
### Step 2: Find intercepts
- [tex]\( x \)[/tex]-intercepts:
- For [tex]\( x \geq 0 \)[/tex], we set [tex]\( g(x) = -x^2 \)[/tex] and solve for [tex]\( x \)[/tex].
[tex]\[ -x^2 = 0 \][/tex]
[tex]\[ x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
Therefore, there is one [tex]\( x \)[/tex]-intercept at [tex]\( x = 0 \)[/tex].
- [tex]\( y \)[/tex]-intercepts:
- The [tex]\( y \)[/tex]-intercept is the value of [tex]\( g \)[/tex] when [tex]\( x = 0 \)[/tex]. For [tex]\( x = 0 \)[/tex], we use the part of the piecewise function that applies to [tex]\( x \geq 0 \)[/tex]:
[tex]\[ g(0) = -0^2 = 0 \][/tex]
Therefore, there is one [tex]\( y \)[/tex]-intercept at [tex]\( y = 0 \)[/tex].
### Step 3: Determine where the function is decreasing
- For [tex]\( x < 0 \)[/tex], [tex]\( g(x) = \left( \frac{3}{4} \right)^x \)[/tex]:
- [tex]\( \left( \frac{3}{4} \right)^x \)[/tex] decreases as [tex]\( x \to -\infty \)[/tex] and increases as [tex]\( x \to 0 \)[/tex], but we consider it increasing when [tex]\( x < 0 \)[/tex].
- For [tex]\( x \geq 0 \)[/tex], [tex]\( g(x) = -x^2 \)[/tex]:
- [tex]\( -x^2 \)[/tex] is a decreasing function for [tex]\( x \geq 0 \)[/tex].
Combining these observations, [tex]\( g(x) \)[/tex] is decreasing for [tex]\( x \geq 0 \)[/tex].
### Step 4: Continuity
- To check continuity, the function must be continuous at the point where the pieces of the function meet, which is [tex]\( x = 0 \)[/tex].
- As [tex]\( x \to 0^- \)[/tex], the value of [tex]\( g(x) = \left( \frac{3}{4} \right)^x \)[/tex] approaches [tex]\( 1 \)[/tex].
- As [tex]\( x \to 0^+ \)[/tex], the value of [tex]\( g(x) = -x^2 \)[/tex] goes to [tex]\( 0 \)[/tex].
Since there is a jump from 1 to 0 at [tex]\( x = 0 \)[/tex], the function [tex]\( g \)[/tex] is not continuous at [tex]\( x = 0 \)[/tex].
### Step 5: Conclusion
Based on the above analysis:
- The function [tex]\( g \)[/tex] has 1 [tex]\( x \)[/tex]-intercept(s) and 1 [tex]\( y \)[/tex]-intercept(s).
- The function [tex]\( g \)[/tex] is sometimes decreasing.
- The function [tex]\( g \)[/tex] is not continuous.
### Final Result:
So, filling in the blanks we get:
- Function [tex]\( g \)[/tex] has 1 [tex]\( x \)[/tex]-intercept(s) and 1 [tex]\( y \)[/tex]-intercept(s).
- Function [tex]\( g \)[/tex] is sometimes decreasing.
- Function [tex]\( g \)[/tex] not continuous.
### Step 1: Define the function
The function [tex]\( g \)[/tex] is defined piecewise as:
[tex]\[ g(x) = \begin{cases} \left(\frac{3}{4}\right)^x & \text{for } x < 0 \\ -x^2 & \text{for } x \geq 0 \end{cases} \][/tex]
### Step 2: Find intercepts
- [tex]\( x \)[/tex]-intercepts:
- For [tex]\( x \geq 0 \)[/tex], we set [tex]\( g(x) = -x^2 \)[/tex] and solve for [tex]\( x \)[/tex].
[tex]\[ -x^2 = 0 \][/tex]
[tex]\[ x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
Therefore, there is one [tex]\( x \)[/tex]-intercept at [tex]\( x = 0 \)[/tex].
- [tex]\( y \)[/tex]-intercepts:
- The [tex]\( y \)[/tex]-intercept is the value of [tex]\( g \)[/tex] when [tex]\( x = 0 \)[/tex]. For [tex]\( x = 0 \)[/tex], we use the part of the piecewise function that applies to [tex]\( x \geq 0 \)[/tex]:
[tex]\[ g(0) = -0^2 = 0 \][/tex]
Therefore, there is one [tex]\( y \)[/tex]-intercept at [tex]\( y = 0 \)[/tex].
### Step 3: Determine where the function is decreasing
- For [tex]\( x < 0 \)[/tex], [tex]\( g(x) = \left( \frac{3}{4} \right)^x \)[/tex]:
- [tex]\( \left( \frac{3}{4} \right)^x \)[/tex] decreases as [tex]\( x \to -\infty \)[/tex] and increases as [tex]\( x \to 0 \)[/tex], but we consider it increasing when [tex]\( x < 0 \)[/tex].
- For [tex]\( x \geq 0 \)[/tex], [tex]\( g(x) = -x^2 \)[/tex]:
- [tex]\( -x^2 \)[/tex] is a decreasing function for [tex]\( x \geq 0 \)[/tex].
Combining these observations, [tex]\( g(x) \)[/tex] is decreasing for [tex]\( x \geq 0 \)[/tex].
### Step 4: Continuity
- To check continuity, the function must be continuous at the point where the pieces of the function meet, which is [tex]\( x = 0 \)[/tex].
- As [tex]\( x \to 0^- \)[/tex], the value of [tex]\( g(x) = \left( \frac{3}{4} \right)^x \)[/tex] approaches [tex]\( 1 \)[/tex].
- As [tex]\( x \to 0^+ \)[/tex], the value of [tex]\( g(x) = -x^2 \)[/tex] goes to [tex]\( 0 \)[/tex].
Since there is a jump from 1 to 0 at [tex]\( x = 0 \)[/tex], the function [tex]\( g \)[/tex] is not continuous at [tex]\( x = 0 \)[/tex].
### Step 5: Conclusion
Based on the above analysis:
- The function [tex]\( g \)[/tex] has 1 [tex]\( x \)[/tex]-intercept(s) and 1 [tex]\( y \)[/tex]-intercept(s).
- The function [tex]\( g \)[/tex] is sometimes decreasing.
- The function [tex]\( g \)[/tex] is not continuous.
### Final Result:
So, filling in the blanks we get:
- Function [tex]\( g \)[/tex] has 1 [tex]\( x \)[/tex]-intercept(s) and 1 [tex]\( y \)[/tex]-intercept(s).
- Function [tex]\( g \)[/tex] is sometimes decreasing.
- Function [tex]\( g \)[/tex] not continuous.
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.