Answer :
To find the intervals on which the function [tex]\( f(x) = 2x^4 + 16x^3 - 21 \)[/tex] is decreasing, we need to follow these steps:
1. Find the derivative:
- First, calculate the derivative of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex].
- The derivative, [tex]\( f'(x) \)[/tex], gives us the rate at which [tex]\( f(x) \)[/tex] is changing.
- For the function [tex]\( f(x) = 2x^4 + 16x^3 - 21 \)[/tex], the derivative is:
[tex]\[
f'(x) = 8x^3 + 48x^2
\][/tex]
2. Find critical points:
- Critical points occur where the derivative is zero or undefined. Here, we set [tex]\( f'(x) = 0 \)[/tex] to find the critical points.
- Solving [tex]\( 8x^3 + 48x^2 = 0 \)[/tex], factoring out the common term [tex]\( 8x^2 \)[/tex], gives:
[tex]\[
8x^2(x + 6) = 0
\][/tex]
- This results in two solutions: [tex]\( x = 0 \)[/tex] and [tex]\( x = -6 \)[/tex].
3. Determine intervals of increase and decrease:
- Use the critical points to divide the number line into intervals.
- You will have the intervals: [tex]\( (-\infty, -6) \)[/tex], [tex]\( (-6, 0) \)[/tex], and [tex]\( (0, \infty) \)[/tex].
- For each interval, test the sign of [tex]\( f'(x) \)[/tex] to determine if the function is increasing or decreasing:
- Interval [tex]\((- \infty, -6)\)[/tex]:
- Choose a number in this interval, such as [tex]\( x = -7 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( 8(-7)^3 + 48(-7)^2 \)[/tex].
- The result is positive, so the function is not decreasing in this interval.
- Interval [tex]\((-6, 0)\)[/tex]:
- Choose a number in this interval, such as [tex]\( x = -1 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( 8(-1)^3 + 48(-1)^2 \)[/tex].
- The result is negative, indicating the function is decreasing in this interval.
- Interval [tex]\((0, \infty)\)[/tex]:
- Choose a number in this interval, such as [tex]\( x = 1 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( 8(1)^3 + 48(1)^2 \)[/tex].
- The result is positive, so the function is not decreasing in this interval.
4. Conclusion:
- The function [tex]\( f(x) = 2x^4 + 16x^3 - 21 \)[/tex] is decreasing on the interval [tex]\( (-6, 0) \)[/tex].
1. Find the derivative:
- First, calculate the derivative of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex].
- The derivative, [tex]\( f'(x) \)[/tex], gives us the rate at which [tex]\( f(x) \)[/tex] is changing.
- For the function [tex]\( f(x) = 2x^4 + 16x^3 - 21 \)[/tex], the derivative is:
[tex]\[
f'(x) = 8x^3 + 48x^2
\][/tex]
2. Find critical points:
- Critical points occur where the derivative is zero or undefined. Here, we set [tex]\( f'(x) = 0 \)[/tex] to find the critical points.
- Solving [tex]\( 8x^3 + 48x^2 = 0 \)[/tex], factoring out the common term [tex]\( 8x^2 \)[/tex], gives:
[tex]\[
8x^2(x + 6) = 0
\][/tex]
- This results in two solutions: [tex]\( x = 0 \)[/tex] and [tex]\( x = -6 \)[/tex].
3. Determine intervals of increase and decrease:
- Use the critical points to divide the number line into intervals.
- You will have the intervals: [tex]\( (-\infty, -6) \)[/tex], [tex]\( (-6, 0) \)[/tex], and [tex]\( (0, \infty) \)[/tex].
- For each interval, test the sign of [tex]\( f'(x) \)[/tex] to determine if the function is increasing or decreasing:
- Interval [tex]\((- \infty, -6)\)[/tex]:
- Choose a number in this interval, such as [tex]\( x = -7 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( 8(-7)^3 + 48(-7)^2 \)[/tex].
- The result is positive, so the function is not decreasing in this interval.
- Interval [tex]\((-6, 0)\)[/tex]:
- Choose a number in this interval, such as [tex]\( x = -1 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( 8(-1)^3 + 48(-1)^2 \)[/tex].
- The result is negative, indicating the function is decreasing in this interval.
- Interval [tex]\((0, \infty)\)[/tex]:
- Choose a number in this interval, such as [tex]\( x = 1 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( 8(1)^3 + 48(1)^2 \)[/tex].
- The result is positive, so the function is not decreasing in this interval.
4. Conclusion:
- The function [tex]\( f(x) = 2x^4 + 16x^3 - 21 \)[/tex] is decreasing on the interval [tex]\( (-6, 0) \)[/tex].
