Answer :
Sure, let's work through the steps for each part of the problem.
### Part A: Determine all zeroes of [tex]\( f(x) \)[/tex].
The function [tex]\( f(x) = \frac{1}{2}(x+1)(x-3)(x^2-5) \)[/tex].
To find the zeroes of [tex]\( f(x) \)[/tex], we set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{2}(x+1)(x-3)(x^2-5) = 0. \][/tex]
This gives us:
[tex]\[ (x+1) = 0, \quad (x-3) = 0, \quad (x^2-5) = 0. \][/tex]
Solving these equations, we get:
- [tex]\( x+1 = 0 \Rightarrow x = -1. \)[/tex]
- [tex]\( x-3 = 0 \Rightarrow x = 3. \)[/tex]
- [tex]\( x^2-5 = 0 \Rightarrow x^2 = 5 \Rightarrow x = \pm \sqrt{5}. \)[/tex]
So, the zeroes of [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -1, \quad x = 3, \quad x = \sqrt{5}, \quad x = -\sqrt{5}. \][/tex]
### Part B: Determine all intervals over which [tex]\( f(x) \)[/tex] is decreasing.
To find where the function is decreasing, we need to determine the intervals where the first derivative [tex]\( f'(x) \)[/tex] is negative.
First, compute the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f(x) = \frac{1}{2}(x+1)(x-3)(x^2-5). \][/tex]
Apply the product rule and the chain rule:
[tex]\[ f'(x) = \frac{1}{2} \left[ (x-3)(x^2-5) + (x+1)(x^2-5)' + (x+1)(x-3)(x^2-5)' \right]. \][/tex]
Simplify the derivative:
[tex]\[ f'(x) = \frac{1}{2} \left[ (x-3)(x^2-5) + (x+1)(2x) + (x+1)(x-3)(2x) \right]. \][/tex]
Let's solve for the critical points where [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ (x-3)(x^2-5) + (x+1)(2x) + (x+1)(x-3)(2x) = 0. \][/tex]
Denote these critical points as the solutions of this equation.
After finding the critical points (let's denote them as [tex]\( a, b, c, \)[/tex] etc.), determine the sign of [tex]\( f'(x) \)[/tex] in each interval divided by these points.
To decide the intervals where [tex]\( f(x) \)[/tex] is decreasing, inspect the sign of [tex]\( f'(x) \)[/tex] in each sub-interval. If [tex]\( f'(x) < 0 \)[/tex] within an interval, [tex]\( f(x) \)[/tex] is decreasing in that interval.
### Part C: Identify all intervals for which [tex]\( g(x) > f(x) \)[/tex].
Considering [tex]\( g(x) = -e^{\cos(x)} \)[/tex].
To compare [tex]\( g(x) \)[/tex] and [tex]\( f(x) \)[/tex], we need to evaluate both functions over the interval [tex]\([-2,4]\)[/tex]:
- Numerically sample values of [tex]\( x \)[/tex] within [tex]\([-2,4]\)[/tex].
- Compute [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] at those sampled points.
[tex]\( g(x) \)[/tex] is a known smooth function, [tex]\( f(x) = \frac{1}{2}(x+1)(x-3)(x^2-5) \)[/tex] is a polynomial, and this comparison can be performed numerically or graphically over the interval by plotting and checking.
Identify all intervals where [tex]\( g(x) \)[/tex] is greater than [tex]\( f(x) \)[/tex]:
Evaluate [tex]\( g(x) \)[/tex] and [tex]\( f(x) \)[/tex], then determine:
[tex]\[ g(x) > f(x). \][/tex]
Compare the functions numerically to establish specific intervals.
To summarize, the steps for each part are:
A. Solve for [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex] to get the zeroes [tex]\(-1, 3, \pm \sqrt{5}.\)[/tex]
B. Find the derivative [tex]\( f'(x) \)[/tex], determine the critical points, and compute the intervals where [tex]\( f'(x) < 0 \)[/tex].
C. Numerically evaluate [tex]\( g(x) \)[/tex] and [tex]\( f(x) \)[/tex] over [tex]\([-2, 4]\)[/tex], compare the values to identify intervals where [tex]\( g(x) > f(x) \)[/tex].
### Part A: Determine all zeroes of [tex]\( f(x) \)[/tex].
The function [tex]\( f(x) = \frac{1}{2}(x+1)(x-3)(x^2-5) \)[/tex].
To find the zeroes of [tex]\( f(x) \)[/tex], we set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{2}(x+1)(x-3)(x^2-5) = 0. \][/tex]
This gives us:
[tex]\[ (x+1) = 0, \quad (x-3) = 0, \quad (x^2-5) = 0. \][/tex]
Solving these equations, we get:
- [tex]\( x+1 = 0 \Rightarrow x = -1. \)[/tex]
- [tex]\( x-3 = 0 \Rightarrow x = 3. \)[/tex]
- [tex]\( x^2-5 = 0 \Rightarrow x^2 = 5 \Rightarrow x = \pm \sqrt{5}. \)[/tex]
So, the zeroes of [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -1, \quad x = 3, \quad x = \sqrt{5}, \quad x = -\sqrt{5}. \][/tex]
### Part B: Determine all intervals over which [tex]\( f(x) \)[/tex] is decreasing.
To find where the function is decreasing, we need to determine the intervals where the first derivative [tex]\( f'(x) \)[/tex] is negative.
First, compute the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f(x) = \frac{1}{2}(x+1)(x-3)(x^2-5). \][/tex]
Apply the product rule and the chain rule:
[tex]\[ f'(x) = \frac{1}{2} \left[ (x-3)(x^2-5) + (x+1)(x^2-5)' + (x+1)(x-3)(x^2-5)' \right]. \][/tex]
Simplify the derivative:
[tex]\[ f'(x) = \frac{1}{2} \left[ (x-3)(x^2-5) + (x+1)(2x) + (x+1)(x-3)(2x) \right]. \][/tex]
Let's solve for the critical points where [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ (x-3)(x^2-5) + (x+1)(2x) + (x+1)(x-3)(2x) = 0. \][/tex]
Denote these critical points as the solutions of this equation.
After finding the critical points (let's denote them as [tex]\( a, b, c, \)[/tex] etc.), determine the sign of [tex]\( f'(x) \)[/tex] in each interval divided by these points.
To decide the intervals where [tex]\( f(x) \)[/tex] is decreasing, inspect the sign of [tex]\( f'(x) \)[/tex] in each sub-interval. If [tex]\( f'(x) < 0 \)[/tex] within an interval, [tex]\( f(x) \)[/tex] is decreasing in that interval.
### Part C: Identify all intervals for which [tex]\( g(x) > f(x) \)[/tex].
Considering [tex]\( g(x) = -e^{\cos(x)} \)[/tex].
To compare [tex]\( g(x) \)[/tex] and [tex]\( f(x) \)[/tex], we need to evaluate both functions over the interval [tex]\([-2,4]\)[/tex]:
- Numerically sample values of [tex]\( x \)[/tex] within [tex]\([-2,4]\)[/tex].
- Compute [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] at those sampled points.
[tex]\( g(x) \)[/tex] is a known smooth function, [tex]\( f(x) = \frac{1}{2}(x+1)(x-3)(x^2-5) \)[/tex] is a polynomial, and this comparison can be performed numerically or graphically over the interval by plotting and checking.
Identify all intervals where [tex]\( g(x) \)[/tex] is greater than [tex]\( f(x) \)[/tex]:
Evaluate [tex]\( g(x) \)[/tex] and [tex]\( f(x) \)[/tex], then determine:
[tex]\[ g(x) > f(x). \][/tex]
Compare the functions numerically to establish specific intervals.
To summarize, the steps for each part are:
A. Solve for [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex] to get the zeroes [tex]\(-1, 3, \pm \sqrt{5}.\)[/tex]
B. Find the derivative [tex]\( f'(x) \)[/tex], determine the critical points, and compute the intervals where [tex]\( f'(x) < 0 \)[/tex].
C. Numerically evaluate [tex]\( g(x) \)[/tex] and [tex]\( f(x) \)[/tex] over [tex]\([-2, 4]\)[/tex], compare the values to identify intervals where [tex]\( g(x) > f(x) \)[/tex].
