Get expert advice and community support for your questions on IDNLearn.com. Our experts are ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
Let's solve the problem step by step.
### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
We are given the function [tex]\( f(x) = -\frac{7}{-3 + x^4} \)[/tex]. We need to find its derivative [tex]\( f'(x) \)[/tex].
Using the quotient rule:
[tex]\[ f'(x) = \frac{d}{dx} \left( -\frac{7}{-3 + x^4} \right) \][/tex]
### Step 2: Apply the Quotient Rule
The quotient rule states that for a function [tex]\( g(x) = \frac{u(x)}{v(x)} \)[/tex], the derivative is given by:
[tex]\[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \][/tex]
For [tex]\( f(x) = u(x) / v(x) \)[/tex] where [tex]\( u(x) = -7 \)[/tex] and [tex]\( v(x) = -3 + x^4 \)[/tex]:
- [tex]\( u'(x) = 0 \)[/tex] (since [tex]\( u(x) = -7 \)[/tex] is a constant)
- [tex]\( v'(x) = 4x^3 \)[/tex]
Using the quotient rule, we get:
[tex]\[ f'(x) = \frac{0 \cdot (-3 + x^4) - (-7) \cdot 4x^3}{(-3 + x^4)^2} = \frac{28 x^3}{(x^4 - 3)^2} \][/tex]
Hence, the derivative is:
[tex]\[ f'(x) = \frac{28 x^3}{(x^4 - 3)^2} \][/tex]
### Step 3: Determine where the denominator is zero
The next step is to find the singularities of [tex]\( f'(x) \)[/tex], i.e., the values of [tex]\( x \)[/tex] where the denominator [tex]\( (x^4 - 3)^2 \)[/tex] is zero.
Solve for [tex]\( x \)[/tex]:
[tex]\[ x^4 - 3 = 0 \implies x^4 = 3 \implies x = \pm \sqrt[4]{3} \text{ or } x = \pm i \sqrt[4]{3} \][/tex]
So, the roots (or singularities) are:
[tex]\[ x = \sqrt[4]{3},\; x = -\sqrt[4]{3},\; x = i \sqrt[4]{3},\; x = -i \sqrt[4]{3} \][/tex]
### Step 4: Determine the radius of convergence
The interval of convergence is determined by the distance to the nearest singularity from the center of the power series (in this case centered at [tex]\( 0 \)[/tex]).
The distance from [tex]\( 0 \)[/tex] to each of these roots is [tex]\( \sqrt[4]{3} \)[/tex]. Thus, the radius of convergence [tex]\( R \)[/tex] is [tex]\( \sqrt[4]{3} \)[/tex].
### Step 5: Express the interval of convergence
The interval of convergence is given as [tex]\( (-R, R) \)[/tex]:
[tex]\[ (-\sqrt[4]{3}, \sqrt[4]{3}) \][/tex]
Hence, the exact interval of convergence for [tex]\( f'(x) \)[/tex] is:
[tex]\[ (-\sqrt[4]{3}, \sqrt[4]{3}) \][/tex]
### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
We are given the function [tex]\( f(x) = -\frac{7}{-3 + x^4} \)[/tex]. We need to find its derivative [tex]\( f'(x) \)[/tex].
Using the quotient rule:
[tex]\[ f'(x) = \frac{d}{dx} \left( -\frac{7}{-3 + x^4} \right) \][/tex]
### Step 2: Apply the Quotient Rule
The quotient rule states that for a function [tex]\( g(x) = \frac{u(x)}{v(x)} \)[/tex], the derivative is given by:
[tex]\[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \][/tex]
For [tex]\( f(x) = u(x) / v(x) \)[/tex] where [tex]\( u(x) = -7 \)[/tex] and [tex]\( v(x) = -3 + x^4 \)[/tex]:
- [tex]\( u'(x) = 0 \)[/tex] (since [tex]\( u(x) = -7 \)[/tex] is a constant)
- [tex]\( v'(x) = 4x^3 \)[/tex]
Using the quotient rule, we get:
[tex]\[ f'(x) = \frac{0 \cdot (-3 + x^4) - (-7) \cdot 4x^3}{(-3 + x^4)^2} = \frac{28 x^3}{(x^4 - 3)^2} \][/tex]
Hence, the derivative is:
[tex]\[ f'(x) = \frac{28 x^3}{(x^4 - 3)^2} \][/tex]
### Step 3: Determine where the denominator is zero
The next step is to find the singularities of [tex]\( f'(x) \)[/tex], i.e., the values of [tex]\( x \)[/tex] where the denominator [tex]\( (x^4 - 3)^2 \)[/tex] is zero.
Solve for [tex]\( x \)[/tex]:
[tex]\[ x^4 - 3 = 0 \implies x^4 = 3 \implies x = \pm \sqrt[4]{3} \text{ or } x = \pm i \sqrt[4]{3} \][/tex]
So, the roots (or singularities) are:
[tex]\[ x = \sqrt[4]{3},\; x = -\sqrt[4]{3},\; x = i \sqrt[4]{3},\; x = -i \sqrt[4]{3} \][/tex]
### Step 4: Determine the radius of convergence
The interval of convergence is determined by the distance to the nearest singularity from the center of the power series (in this case centered at [tex]\( 0 \)[/tex]).
The distance from [tex]\( 0 \)[/tex] to each of these roots is [tex]\( \sqrt[4]{3} \)[/tex]. Thus, the radius of convergence [tex]\( R \)[/tex] is [tex]\( \sqrt[4]{3} \)[/tex].
### Step 5: Express the interval of convergence
The interval of convergence is given as [tex]\( (-R, R) \)[/tex]:
[tex]\[ (-\sqrt[4]{3}, \sqrt[4]{3}) \][/tex]
Hence, the exact interval of convergence for [tex]\( f'(x) \)[/tex] is:
[tex]\[ (-\sqrt[4]{3}, \sqrt[4]{3}) \][/tex]
We greatly 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. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.