Answer :
Let's go through each part step-by-step for the function [tex]\( f(x) = 2x - 3x^{\frac{2}{3}} \)[/tex].
### (a) Domain
The domain of [tex]\( f(x) \)[/tex] is determined by where the function is defined. Since the function involves [tex]\( 2x \)[/tex] and [tex]\( 3x^{\frac{2}{3}} \)[/tex], both terms are defined for all real [tex]\( x \)[/tex]. Therefore, the domain is:
[tex]\[ (-\infty, \infty) \][/tex]
### (b) Asymptotes
#### Vertical Asymptotes
Vertical asymptotes occur where the function goes to [tex]\(\pm \infty\)[/tex], typically due to dividing by zero or similar problems. Since the function [tex]\( f(x) = 2x - 3x^{\frac{2}{3}} \)[/tex] is defined for all [tex]\( x \)[/tex], there are no vertical asymptotes.
#### Horizontal Asymptotes
Horizontal asymptotes are determined by the end behavior of the function as [tex]\( x \)[/tex] approaches [tex]\(\pm\infty\)[/tex]. To find these:
For [tex]\( x \to \infty \)[/tex]:
[tex]\[ f(x) = 2x - 3x^{\frac{2}{3}} \][/tex]
As [tex]\( x \)[/tex] becomes very large, the term [tex]\( 2x \)[/tex] will dominate [tex]\( 3x^{\frac{2}{3}} \)[/tex]. Therefore, as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex]. Similarly, as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]. Thus, there are no horizontal asymptotes.
### (c) Symmetry
To determine the symmetry, we check [tex]\( f(-x) \)[/tex] and compare it to [tex]\( f(x) \)[/tex].
[tex]\[ f(-x) = 2(-x) - 3(-x)^{\frac{2}{3}} = -2x - 3x^{\frac{2}{3}} \][/tex]
Since [tex]\( f(-x) \neq f(x) \)[/tex] and [tex]\( f(-x) \neq -f(x) \)[/tex], the function is neither even nor odd.
### (d) Increasing / Decreasing
To determine where the function is increasing or decreasing, we need to find the first derivative [tex]\( f'(x) \)[/tex].
[tex]\[ f'(x) = \frac{d}{dx} \left( 2x - 3x^{\frac{2}{3}} \right) = 2 - 2x^{-\frac{1}{3}} \][/tex]
Set [tex]\( f'(x) = 0 \)[/tex] to find critical points:
[tex]\[ 2 - 2x^{-\frac{1}{3}} = 0 \implies 2 = 2x^{-\frac{1}{3}} \implies x^{-\frac{1}{3}} = 1 \implies x = 1 \][/tex]
#### Increasing / Decreasing Intervals
To determine the sign of [tex]\( f'(x) \)[/tex]:
1. For [tex]\( x < 1 \)[/tex], [tex]\( f'(x) \)[/tex] is positive (since [tex]\( x^{-\frac{1}{3}} > 1 \)[/tex]).
2. For [tex]\( x > 1 \)[/tex], [tex]\( f'(x) \)[/tex] is negative (since [tex]\( x^{-\frac{1}{3}} < 1 \)[/tex]).
So, [tex]\( f \)[/tex] is:
- Increasing for [tex]\( x \in (-\infty, 1) \)[/tex]
- Decreasing for [tex]\( x \in (1, \infty) \)[/tex]
### (d) Critical Point Classification
To classify the critical point at [tex]\( x = 1 \)[/tex], we examine the second derivative [tex]\( f''(x) \)[/tex].
[tex]\[ f''(x) = \frac{d}{dx} \left( 2 - 2x^{-\frac{1}{3}} \right) = 2 \cdot \left(-\frac{1}{3}\right) x^{-\frac{4}{3}} = -\frac{2}{3} x^{-\frac{4}{3}} \][/tex]
Evaluate [tex]\( f''(x) \)[/tex] at [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = -\frac{2}{3} \][/tex]
Since [tex]\( f''(1) < 0 \)[/tex], the function has a local maximum at [tex]\( x = 1 \)[/tex].
### (e) Concavity
The second derivative [tex]\( f''(x) = -\frac{2}{3} x^{-\frac{4}{3}} \)[/tex] tells us about concavity:
- Since [tex]\( f''(x) < 0 \)[/tex] for all [tex]\( x \neq 0 \)[/tex], the function is concave down everywhere except at [tex]\( x = 0 \)[/tex].
### (f) Inflection Points
Inflection points occur where [tex]\( f''(x) \)[/tex] changes sign. However, [tex]\( f''(x) \)[/tex] does not change sign (it is always negative for all [tex]\( x \neq 0 \)[/tex]). So, there are no inflection points.
Hence:
(d) Increasing / Decreasing Intervals:
- [tex]\( f \)[/tex] is increasing for [tex]\( x \in (-\infty, 1) \)[/tex]
- [tex]\( f \)[/tex] is decreasing for [tex]\( x \in (1, \infty) \)[/tex]
(d) Critical Point Classification:
- [tex]\( f \)[/tex] has a local maximum at [tex]\( x = 1 \)[/tex]
- [tex]\( f \)[/tex] has no local minimums
- [tex]\( f \)[/tex] has critical points that are neither local mins nor maxes at [tex]\( x = \)[/tex] NONE
(e) Concavity:
- [tex]\( f \)[/tex] is concave up for [tex]\( x \in \)[/tex] NONE
- [tex]\( f \)[/tex] is concave down for [tex]\( x \in (-\infty, 0) \cup (0, \infty) \)[/tex]
(f) Inflection Points:
- [tex]\( f \)[/tex] has inflection points at [tex]\( x = \)[/tex] NONE
### (a) Domain
The domain of [tex]\( f(x) \)[/tex] is determined by where the function is defined. Since the function involves [tex]\( 2x \)[/tex] and [tex]\( 3x^{\frac{2}{3}} \)[/tex], both terms are defined for all real [tex]\( x \)[/tex]. Therefore, the domain is:
[tex]\[ (-\infty, \infty) \][/tex]
### (b) Asymptotes
#### Vertical Asymptotes
Vertical asymptotes occur where the function goes to [tex]\(\pm \infty\)[/tex], typically due to dividing by zero or similar problems. Since the function [tex]\( f(x) = 2x - 3x^{\frac{2}{3}} \)[/tex] is defined for all [tex]\( x \)[/tex], there are no vertical asymptotes.
#### Horizontal Asymptotes
Horizontal asymptotes are determined by the end behavior of the function as [tex]\( x \)[/tex] approaches [tex]\(\pm\infty\)[/tex]. To find these:
For [tex]\( x \to \infty \)[/tex]:
[tex]\[ f(x) = 2x - 3x^{\frac{2}{3}} \][/tex]
As [tex]\( x \)[/tex] becomes very large, the term [tex]\( 2x \)[/tex] will dominate [tex]\( 3x^{\frac{2}{3}} \)[/tex]. Therefore, as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex]. Similarly, as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex]. Thus, there are no horizontal asymptotes.
### (c) Symmetry
To determine the symmetry, we check [tex]\( f(-x) \)[/tex] and compare it to [tex]\( f(x) \)[/tex].
[tex]\[ f(-x) = 2(-x) - 3(-x)^{\frac{2}{3}} = -2x - 3x^{\frac{2}{3}} \][/tex]
Since [tex]\( f(-x) \neq f(x) \)[/tex] and [tex]\( f(-x) \neq -f(x) \)[/tex], the function is neither even nor odd.
### (d) Increasing / Decreasing
To determine where the function is increasing or decreasing, we need to find the first derivative [tex]\( f'(x) \)[/tex].
[tex]\[ f'(x) = \frac{d}{dx} \left( 2x - 3x^{\frac{2}{3}} \right) = 2 - 2x^{-\frac{1}{3}} \][/tex]
Set [tex]\( f'(x) = 0 \)[/tex] to find critical points:
[tex]\[ 2 - 2x^{-\frac{1}{3}} = 0 \implies 2 = 2x^{-\frac{1}{3}} \implies x^{-\frac{1}{3}} = 1 \implies x = 1 \][/tex]
#### Increasing / Decreasing Intervals
To determine the sign of [tex]\( f'(x) \)[/tex]:
1. For [tex]\( x < 1 \)[/tex], [tex]\( f'(x) \)[/tex] is positive (since [tex]\( x^{-\frac{1}{3}} > 1 \)[/tex]).
2. For [tex]\( x > 1 \)[/tex], [tex]\( f'(x) \)[/tex] is negative (since [tex]\( x^{-\frac{1}{3}} < 1 \)[/tex]).
So, [tex]\( f \)[/tex] is:
- Increasing for [tex]\( x \in (-\infty, 1) \)[/tex]
- Decreasing for [tex]\( x \in (1, \infty) \)[/tex]
### (d) Critical Point Classification
To classify the critical point at [tex]\( x = 1 \)[/tex], we examine the second derivative [tex]\( f''(x) \)[/tex].
[tex]\[ f''(x) = \frac{d}{dx} \left( 2 - 2x^{-\frac{1}{3}} \right) = 2 \cdot \left(-\frac{1}{3}\right) x^{-\frac{4}{3}} = -\frac{2}{3} x^{-\frac{4}{3}} \][/tex]
Evaluate [tex]\( f''(x) \)[/tex] at [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = -\frac{2}{3} \][/tex]
Since [tex]\( f''(1) < 0 \)[/tex], the function has a local maximum at [tex]\( x = 1 \)[/tex].
### (e) Concavity
The second derivative [tex]\( f''(x) = -\frac{2}{3} x^{-\frac{4}{3}} \)[/tex] tells us about concavity:
- Since [tex]\( f''(x) < 0 \)[/tex] for all [tex]\( x \neq 0 \)[/tex], the function is concave down everywhere except at [tex]\( x = 0 \)[/tex].
### (f) Inflection Points
Inflection points occur where [tex]\( f''(x) \)[/tex] changes sign. However, [tex]\( f''(x) \)[/tex] does not change sign (it is always negative for all [tex]\( x \neq 0 \)[/tex]). So, there are no inflection points.
Hence:
(d) Increasing / Decreasing Intervals:
- [tex]\( f \)[/tex] is increasing for [tex]\( x \in (-\infty, 1) \)[/tex]
- [tex]\( f \)[/tex] is decreasing for [tex]\( x \in (1, \infty) \)[/tex]
(d) Critical Point Classification:
- [tex]\( f \)[/tex] has a local maximum at [tex]\( x = 1 \)[/tex]
- [tex]\( f \)[/tex] has no local minimums
- [tex]\( f \)[/tex] has critical points that are neither local mins nor maxes at [tex]\( x = \)[/tex] NONE
(e) Concavity:
- [tex]\( f \)[/tex] is concave up for [tex]\( x \in \)[/tex] NONE
- [tex]\( f \)[/tex] is concave down for [tex]\( x \in (-\infty, 0) \cup (0, \infty) \)[/tex]
(f) Inflection Points:
- [tex]\( f \)[/tex] has inflection points at [tex]\( x = \)[/tex] NONE
