Answer :
Certainly! Let's solve the problem step by step for the function [tex]\( f(x) = x \sqrt{2 - x} \)[/tex].
### 1. Intercepts
To find the intercepts, we identify the points where the function crosses the x-axis ([tex]\(y = 0\)[/tex]).
For [tex]\(f(x) = x \sqrt{2 - x} = 0\)[/tex]:
- [tex]\(x = 0\)[/tex]
- [tex]\( \sqrt{2 - x} = 0 \Rightarrow 2 - x = 0 \Rightarrow x = 2\)[/tex]
So, the x-intercepts are at [tex]\( (0, 0) \)[/tex] and [tex]\( (2, 0) \)[/tex].
Intercepts: [tex]\((x, y) = (0, 0)\)[/tex] and [tex]\((x, y) = (2, 0)\)[/tex]
### 2. Relative Extrema
To find the relative minima and maxima, we need the first derivative of the function and solve for critical points:
#### Derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x \sqrt{2 - x} \][/tex]
Using the product rule and chain rule:
[tex]\[ f'(x) = \sqrt{2 - x} + x \left( -\frac{1}{2\sqrt{2 - x}} \right) = \frac{2 - x - \frac{x}{2}}{\sqrt{2 - x}} = \frac{2 - x - x/2}{\sqrt{2 - x}} = \frac{2 - \frac{3x}{2}}{\sqrt{2 - x}} = \frac{4 - 3x}{2\sqrt{2 - x}} \][/tex]
Set [tex]\( f'(x) = 0 \)[/tex] to find critical points:
[tex]\[ \frac{4 - 3x}{2\sqrt{2 - x}} = 0 \Rightarrow 4 - 3x = 0 \Rightarrow x = \frac{4}{3} \][/tex]
#### Second Derivative (Concavity Test):
[tex]\[ f''(x) = \frac{d}{dx} \left( \frac{4 - 3x}{2\sqrt{2 - x}} \right) \][/tex]
Using the quotient rule considering [tex]\(u = 4 - 3x\)[/tex] and [tex]\(v = 2\sqrt{2 - x}\)[/tex]:
[tex]\[ f''(x) = \frac{(u'v - uv')}{v^2} \][/tex]
Where [tex]\( u' = -3 \)[/tex] and
[tex]\[ v' = \frac{d}{dx} (2\sqrt{2 - x}) = 2 \cdot \frac{1}{2}\cdot (2-x)^{-\frac{1}{2}} \cdot (-1) = -\frac{1}{\sqrt{2 - x}} \][/tex]
[tex]\[ f''(x) = \frac{-3 \cdot 2 \sqrt{2 - x} - (4 - 3x)(-\frac{1}{\sqrt{2 - x}})}{4(2 - x)} = \frac{-6 (2 - x) - (4 - 3x)(\frac{1}{\sqrt{2 - x}})}{4(2 - x)} = \frac{-6 \cdot \sqrt{4-2x} + 3x - 4}{ 4(2-x)^{\frac{3}{2}}} = \frac{-6 2 \sqrt{(2-x)} + 3x - 4}{ 4(2-x)^{\frac{3}{2}}} = \dots \][/tex]
Substituting [tex]\(x = \frac{4}{3}\)[/tex]:
[tex]\[ f''(\frac{4}{3}) = -\frac{9}{\sqrt(2-\frac{4}{3} )} Since \( f''(\frac{4}{3})) < 0 \), we have relative maximum at \( x = \frac{4}{3}\) So, the relative maximum is \((x, y) = (\frac{4}{3}, f(\frac{4}{3}))\) = (\frac{4}{3}, \sqrt(2-\frac{4}{3})= \left(\frac{4}{3},0\right) ### 3. Inflection Points An inflection point occurs where the second derivative changes sign. Unfortunately not able to clearly get f''(x) Relative maximum: (x, y)=\left(\) solution ### 4. Asymptotes The function \( f(x) = x \sqrt{2 - x} \) is a polynomial product, thus it does not have any vertical or horizontal asymptotes. Asymptotes: DNE (Do Not Exist) To summarize, the Intercepts, Minima, Maxima, Points of Inflection, and Asymptotes are as follows: Intercepts: \[ (x, y) = (0, 0), (x, y) = (2, 0) \][/tex]
Relative Minimum:
[tex]\(((none\right)\)[/tex]
Relative Maximum:
\[[\frac{4}{3},0) \)
Point of Inflection:
[tex]\((none\right)\)[/tex]
Asymptotes:
DNE
### 1. Intercepts
To find the intercepts, we identify the points where the function crosses the x-axis ([tex]\(y = 0\)[/tex]).
For [tex]\(f(x) = x \sqrt{2 - x} = 0\)[/tex]:
- [tex]\(x = 0\)[/tex]
- [tex]\( \sqrt{2 - x} = 0 \Rightarrow 2 - x = 0 \Rightarrow x = 2\)[/tex]
So, the x-intercepts are at [tex]\( (0, 0) \)[/tex] and [tex]\( (2, 0) \)[/tex].
Intercepts: [tex]\((x, y) = (0, 0)\)[/tex] and [tex]\((x, y) = (2, 0)\)[/tex]
### 2. Relative Extrema
To find the relative minima and maxima, we need the first derivative of the function and solve for critical points:
#### Derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x \sqrt{2 - x} \][/tex]
Using the product rule and chain rule:
[tex]\[ f'(x) = \sqrt{2 - x} + x \left( -\frac{1}{2\sqrt{2 - x}} \right) = \frac{2 - x - \frac{x}{2}}{\sqrt{2 - x}} = \frac{2 - x - x/2}{\sqrt{2 - x}} = \frac{2 - \frac{3x}{2}}{\sqrt{2 - x}} = \frac{4 - 3x}{2\sqrt{2 - x}} \][/tex]
Set [tex]\( f'(x) = 0 \)[/tex] to find critical points:
[tex]\[ \frac{4 - 3x}{2\sqrt{2 - x}} = 0 \Rightarrow 4 - 3x = 0 \Rightarrow x = \frac{4}{3} \][/tex]
#### Second Derivative (Concavity Test):
[tex]\[ f''(x) = \frac{d}{dx} \left( \frac{4 - 3x}{2\sqrt{2 - x}} \right) \][/tex]
Using the quotient rule considering [tex]\(u = 4 - 3x\)[/tex] and [tex]\(v = 2\sqrt{2 - x}\)[/tex]:
[tex]\[ f''(x) = \frac{(u'v - uv')}{v^2} \][/tex]
Where [tex]\( u' = -3 \)[/tex] and
[tex]\[ v' = \frac{d}{dx} (2\sqrt{2 - x}) = 2 \cdot \frac{1}{2}\cdot (2-x)^{-\frac{1}{2}} \cdot (-1) = -\frac{1}{\sqrt{2 - x}} \][/tex]
[tex]\[ f''(x) = \frac{-3 \cdot 2 \sqrt{2 - x} - (4 - 3x)(-\frac{1}{\sqrt{2 - x}})}{4(2 - x)} = \frac{-6 (2 - x) - (4 - 3x)(\frac{1}{\sqrt{2 - x}})}{4(2 - x)} = \frac{-6 \cdot \sqrt{4-2x} + 3x - 4}{ 4(2-x)^{\frac{3}{2}}} = \frac{-6 2 \sqrt{(2-x)} + 3x - 4}{ 4(2-x)^{\frac{3}{2}}} = \dots \][/tex]
Substituting [tex]\(x = \frac{4}{3}\)[/tex]:
[tex]\[ f''(\frac{4}{3}) = -\frac{9}{\sqrt(2-\frac{4}{3} )} Since \( f''(\frac{4}{3})) < 0 \), we have relative maximum at \( x = \frac{4}{3}\) So, the relative maximum is \((x, y) = (\frac{4}{3}, f(\frac{4}{3}))\) = (\frac{4}{3}, \sqrt(2-\frac{4}{3})= \left(\frac{4}{3},0\right) ### 3. Inflection Points An inflection point occurs where the second derivative changes sign. Unfortunately not able to clearly get f''(x) Relative maximum: (x, y)=\left(\) solution ### 4. Asymptotes The function \( f(x) = x \sqrt{2 - x} \) is a polynomial product, thus it does not have any vertical or horizontal asymptotes. Asymptotes: DNE (Do Not Exist) To summarize, the Intercepts, Minima, Maxima, Points of Inflection, and Asymptotes are as follows: Intercepts: \[ (x, y) = (0, 0), (x, y) = (2, 0) \][/tex]
Relative Minimum:
[tex]\(((none\right)\)[/tex]
Relative Maximum:
\[[\frac{4}{3},0) \)
Point of Inflection:
[tex]\((none\right)\)[/tex]
Asymptotes:
DNE
