Answer :
Certainly! Let's analyze the function [tex]\( f(x) = x^4 + 5x^3 - 48x^2 - 252x \)[/tex].
### Finding the Roots
To find the roots of the function [tex]\( f(x) \)[/tex], we need to solve the equation [tex]\( f(x) = 0 \)[/tex]. The roots are the values of [tex]\( x \)[/tex] at which the function equals zero. The roots are:
- [tex]\( x = -6 \)[/tex]
- [tex]\( x = 0 \)[/tex]
- [tex]\( x = 7 \)[/tex]
These roots mean the graph of the function crosses the x-axis at these points.
### Finding the Critical Points
Critical points occur where the derivative of the function equals zero or is undefined. Let's find the derivative of [tex]\( f(x) \)[/tex]:
1. Differentiate the function:
[tex]\[ f'(x) = \frac{d}{dx}(x^4 + 5x^3 - 48x^2 - 252x) \][/tex]
[tex]\[ f'(x) = 4x^3 + 15x^2 - 96x - 252 \][/tex]
2. Set the derivative equal to zero to find the critical points:
[tex]\[ 4x^3 + 15x^2 - 96x - 252 = 0 \][/tex]
Solving this equation gives us the critical points. The critical points are:
- [tex]\( x = -6 \)[/tex]
- [tex]\( x = \frac{9}{8} - \frac{\sqrt{753}}{8} \)[/tex]
- [tex]\( x = \frac{9}{8} + \frac{\sqrt{753}}{8} \)[/tex]
These points give the locations where the function could have local maxima, minima, or points of inflection.
### Conclusion
- The roots of the function are: [tex]\( x = -6 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 7 \)[/tex].
- The critical points are [tex]\( x = -6 \)[/tex], [tex]\( x = \frac{9}{8} - \frac{\sqrt{753}}{8} \)[/tex], and [tex]\( x = \frac{9}{8} + \frac{\sqrt{753}}{8} \)[/tex].
These insights help to understand the behavior of the graph of the function, such as where it intersects the x-axis and where its slope is zero, indicating potential peaks or troughs.
### Finding the Roots
To find the roots of the function [tex]\( f(x) \)[/tex], we need to solve the equation [tex]\( f(x) = 0 \)[/tex]. The roots are the values of [tex]\( x \)[/tex] at which the function equals zero. The roots are:
- [tex]\( x = -6 \)[/tex]
- [tex]\( x = 0 \)[/tex]
- [tex]\( x = 7 \)[/tex]
These roots mean the graph of the function crosses the x-axis at these points.
### Finding the Critical Points
Critical points occur where the derivative of the function equals zero or is undefined. Let's find the derivative of [tex]\( f(x) \)[/tex]:
1. Differentiate the function:
[tex]\[ f'(x) = \frac{d}{dx}(x^4 + 5x^3 - 48x^2 - 252x) \][/tex]
[tex]\[ f'(x) = 4x^3 + 15x^2 - 96x - 252 \][/tex]
2. Set the derivative equal to zero to find the critical points:
[tex]\[ 4x^3 + 15x^2 - 96x - 252 = 0 \][/tex]
Solving this equation gives us the critical points. The critical points are:
- [tex]\( x = -6 \)[/tex]
- [tex]\( x = \frac{9}{8} - \frac{\sqrt{753}}{8} \)[/tex]
- [tex]\( x = \frac{9}{8} + \frac{\sqrt{753}}{8} \)[/tex]
These points give the locations where the function could have local maxima, minima, or points of inflection.
### Conclusion
- The roots of the function are: [tex]\( x = -6 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 7 \)[/tex].
- The critical points are [tex]\( x = -6 \)[/tex], [tex]\( x = \frac{9}{8} - \frac{\sqrt{753}}{8} \)[/tex], and [tex]\( x = \frac{9}{8} + \frac{\sqrt{753}}{8} \)[/tex].
These insights help to understand the behavior of the graph of the function, such as where it intersects the x-axis and where its slope is zero, indicating potential peaks or troughs.
