Answer :
To find the intervals on which the function [tex]\( f(x) = x^4 - 12x^3 - 19 \)[/tex] is decreasing, we need to follow these steps:
1. Find the derivative of the function:
The first step is to find the derivative, [tex]\( f'(x) \)[/tex], which will help us identify the critical points and determine the intervals where the function is increasing or decreasing.
[tex]\[
f'(x) = \frac{d}{dx}(x^4 - 12x^3 - 19) = 4x^3 - 36x^2
\][/tex]
2. Find critical points:
The critical points occur where [tex]\( f'(x) = 0 \)[/tex] or where [tex]\( f'(x) \)[/tex] is undefined (though in this polynomial case, [tex]\( f'(x) \)[/tex] is always defined).
Set the derivative equal to zero:
[tex]\[
4x^3 - 36x^2 = 0
\][/tex]
Factor the equation:
[tex]\[
4x^2(x - 9) = 0
\][/tex]
This gives us critical points at [tex]\( x = 0 \)[/tex] and [tex]\( x = 9 \)[/tex].
3. Determine the sign of [tex]\( f'(x) \)[/tex] around the critical points:
We analyze the sign of [tex]\( f'(x) \)[/tex] in the intervals determined by the critical points: [tex]\( (-\infty, 0) \)[/tex], [tex]\( (0, 9) \)[/tex], and [tex]\( (9, \infty) \)[/tex].
- For [tex]\( (-\infty, 0) \)[/tex]:
Choose a test point, say [tex]\( x = -1 \)[/tex].
[tex]\[
f'(-1) = 4(-1)^3 - 36(-1)^2 = -4 - 36 = -40
\][/tex]
Since [tex]\( f'(-1) < 0 \)[/tex], the function is decreasing on [tex]\( (-\infty, 0) \)[/tex].
- For [tex]\( (0, 9) \)[/tex]:
Choose a test point, say [tex]\( x = 1 \)[/tex].
[tex]\[
f'(1) = 4(1)^3 - 36(1)^2 = 4 - 36 = -32
\][/tex]
Since [tex]\( f'(1) < 0 \)[/tex], the function is also decreasing on [tex]\( (0, 9) \)[/tex].
- For [tex]\( (9, \infty) \)[/tex]:
Choose a test point, say [tex]\( x = 10 \)[/tex].
[tex]\[
f'(10) = 4(10)^3 - 36(10)^2 = 4000 - 3600 = 400
\][/tex]
Since [tex]\( f'(10) > 0 \)[/tex], the function is increasing on [tex]\( (9, \infty) \)[/tex].
4. Conclusion:
The function [tex]\( f(x) \)[/tex] is decreasing on the intervals [tex]\( (-\infty, 0) \)[/tex] and [tex]\( (0, 9) \)[/tex].
So, the intervals on which the function [tex]\( f(x) = x^4 - 12x^3 - 19 \)[/tex] is decreasing are [tex]\( (-\infty, 0) \cup (0, 9) \)[/tex].
1. Find the derivative of the function:
The first step is to find the derivative, [tex]\( f'(x) \)[/tex], which will help us identify the critical points and determine the intervals where the function is increasing or decreasing.
[tex]\[
f'(x) = \frac{d}{dx}(x^4 - 12x^3 - 19) = 4x^3 - 36x^2
\][/tex]
2. Find critical points:
The critical points occur where [tex]\( f'(x) = 0 \)[/tex] or where [tex]\( f'(x) \)[/tex] is undefined (though in this polynomial case, [tex]\( f'(x) \)[/tex] is always defined).
Set the derivative equal to zero:
[tex]\[
4x^3 - 36x^2 = 0
\][/tex]
Factor the equation:
[tex]\[
4x^2(x - 9) = 0
\][/tex]
This gives us critical points at [tex]\( x = 0 \)[/tex] and [tex]\( x = 9 \)[/tex].
3. Determine the sign of [tex]\( f'(x) \)[/tex] around the critical points:
We analyze the sign of [tex]\( f'(x) \)[/tex] in the intervals determined by the critical points: [tex]\( (-\infty, 0) \)[/tex], [tex]\( (0, 9) \)[/tex], and [tex]\( (9, \infty) \)[/tex].
- For [tex]\( (-\infty, 0) \)[/tex]:
Choose a test point, say [tex]\( x = -1 \)[/tex].
[tex]\[
f'(-1) = 4(-1)^3 - 36(-1)^2 = -4 - 36 = -40
\][/tex]
Since [tex]\( f'(-1) < 0 \)[/tex], the function is decreasing on [tex]\( (-\infty, 0) \)[/tex].
- For [tex]\( (0, 9) \)[/tex]:
Choose a test point, say [tex]\( x = 1 \)[/tex].
[tex]\[
f'(1) = 4(1)^3 - 36(1)^2 = 4 - 36 = -32
\][/tex]
Since [tex]\( f'(1) < 0 \)[/tex], the function is also decreasing on [tex]\( (0, 9) \)[/tex].
- For [tex]\( (9, \infty) \)[/tex]:
Choose a test point, say [tex]\( x = 10 \)[/tex].
[tex]\[
f'(10) = 4(10)^3 - 36(10)^2 = 4000 - 3600 = 400
\][/tex]
Since [tex]\( f'(10) > 0 \)[/tex], the function is increasing on [tex]\( (9, \infty) \)[/tex].
4. Conclusion:
The function [tex]\( f(x) \)[/tex] is decreasing on the intervals [tex]\( (-\infty, 0) \)[/tex] and [tex]\( (0, 9) \)[/tex].
So, the intervals on which the function [tex]\( f(x) = x^4 - 12x^3 - 19 \)[/tex] is decreasing are [tex]\( (-\infty, 0) \cup (0, 9) \)[/tex].
