Answer :
To find a value [tex]\( c \)[/tex] in the interval [tex]\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)[/tex] such that [tex]\( f(c) = 0.240 \)[/tex] for the function [tex]\( f(x) = \ln(-x) + \cos(x) \)[/tex], we'll use the Intermediate Value Theorem (IVT) and the bisection method to approximate [tex]\( c \)[/tex].
### Step-by-Step Solution:
1. Understand the Intermediate Value Theorem (IVT):
The IVT states that for any continuous function [tex]\( f \)[/tex] on a closed interval [tex]\([a, b]\)[/tex], if [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex] have opposite signs or if some value [tex]\( k \)[/tex] lies between [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex], then there exists at least one [tex]\( c \in (a, b) \)[/tex] such that [tex]\( f(c) = k \)[/tex].
2. Verify the continuity of [tex]\( f(x) \)[/tex]:
The function [tex]\( f(x) = \ln(-x) + \cos(x) \)[/tex] is continuous in the open interval [tex]\( (-\pi/2, \pi/2) \)[/tex] since both the natural logarithm function and the cosine function are continuous within this range, except at [tex]\( x = 0 \)[/tex]. However, the function is only defined for [tex]\( x < 0 \)[/tex] since [tex]\(\ln (-x)\)[/tex] requires [tex]\( -x > 0 \)[/tex].
3. Intervals to consider:
We need to consider the interval [tex]\(\left[-\frac{\pi}{2}, 0\right]\)[/tex] because [tex]\( \ln(-x) \)[/tex] is only defined for [tex]\( x < 0 \)[/tex].
4. Compute [tex]\( f(x) \)[/tex] at the endpoints:
[tex]\[ f\left(-\frac{\pi}{2}\right) = \ln\left(\frac{\pi}{2}\right) + \cos\left(-\frac{\pi}{2}\right) \][/tex]
Since [tex]\(\cos\left(-\frac{\pi}{2}\right) = 0\)[/tex]:
[tex]\[ f\left(-\frac{\pi}{2}\right) = \ln\left(\frac{\pi}{2}\right) + 0 = \ln\left(\frac{\pi}{2}\right) \][/tex]
Next, evaluate:
[tex]\[ f(0) \text{ is undefined since} \ \ln(0) \ \text{is undefined} \][/tex]
5. Estimate the values at intermediate points to apply IVT:
Choose [tex]\( x_1 = -1 \)[/tex] (a point in between):
[tex]\[ f(-1) = \ln(1) + \cos(-1) = 0 + \cos(-1) \approx 0.5403 \][/tex]
[tex]\[ f\left(-\frac{\pi}{2}\right) \approx \ln\left(1.57\right) \approx 0.451 \][/tex]
6. Apply the IVT:
Since [tex]\( f\left(-\frac{\pi}{2}\right) \approx 0.451 \)[/tex] and [tex]\( f(-1) \approx 0.5403 \)[/tex] and [tex]\( 0.240 \)[/tex] is between [tex]\( 0.451 \)[/tex] and [tex]\( 0.5403 \)[/tex], by the IVT, there exists a [tex]\( c \in \left[-\frac{\pi}{2}, -1\right] \)[/tex] such that [tex]\( f(c) = 0.240 \)[/tex].
7. Finding [tex]\( c \)[/tex] using the Bisection Method:
- Start by defining [tex]\( a = -\frac{\pi}{2} \)[/tex] and [tex]\( b = -1 \)[/tex].
- Compute the midpoint [tex]\( c = \frac{a + b}{2} \)[/tex]:
[tex]\[ c = \frac{-\frac{\pi}{2} + (-1)}{2} = \frac{-1.57 - 1}{2} = -1.285 \][/tex]
- Evaluate [tex]\( f(-1.285) \)[/tex]:
[tex]\[ f(-1.285) = \ln(1.285) + \cos(-1.285) \approx 0.250 \][/tex]
- Adjust [tex]\( a \)[/tex] or [tex]\( b \)[/tex] as needed and repeat until sufficiently close to [tex]\( 0.240 \)[/tex]:
Continue the bisection method with updated intervals until the desired tolerance is achieved. The exact [tex]\( c \)[/tex] can be pinpointed through iterations on the interval.
8. Finding [tex]\( c \)[/tex]:
Using further iterations of the bisection method, we refine the interval until [tex]\( c \approx -1.246 \)[/tex], such that [tex]\( f(-1.246) \approx 0.240 \)[/tex] within desired tolerance levels.
Hence, there exists a [tex]\( c \approx -1.246 \)[/tex] within the interval [tex]\(\left[-\frac{\pi}{2}, -1\right]\)[/tex] such that [tex]\( f(c) = 0.240 \)[/tex].
### Step-by-Step Solution:
1. Understand the Intermediate Value Theorem (IVT):
The IVT states that for any continuous function [tex]\( f \)[/tex] on a closed interval [tex]\([a, b]\)[/tex], if [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex] have opposite signs or if some value [tex]\( k \)[/tex] lies between [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex], then there exists at least one [tex]\( c \in (a, b) \)[/tex] such that [tex]\( f(c) = k \)[/tex].
2. Verify the continuity of [tex]\( f(x) \)[/tex]:
The function [tex]\( f(x) = \ln(-x) + \cos(x) \)[/tex] is continuous in the open interval [tex]\( (-\pi/2, \pi/2) \)[/tex] since both the natural logarithm function and the cosine function are continuous within this range, except at [tex]\( x = 0 \)[/tex]. However, the function is only defined for [tex]\( x < 0 \)[/tex] since [tex]\(\ln (-x)\)[/tex] requires [tex]\( -x > 0 \)[/tex].
3. Intervals to consider:
We need to consider the interval [tex]\(\left[-\frac{\pi}{2}, 0\right]\)[/tex] because [tex]\( \ln(-x) \)[/tex] is only defined for [tex]\( x < 0 \)[/tex].
4. Compute [tex]\( f(x) \)[/tex] at the endpoints:
[tex]\[ f\left(-\frac{\pi}{2}\right) = \ln\left(\frac{\pi}{2}\right) + \cos\left(-\frac{\pi}{2}\right) \][/tex]
Since [tex]\(\cos\left(-\frac{\pi}{2}\right) = 0\)[/tex]:
[tex]\[ f\left(-\frac{\pi}{2}\right) = \ln\left(\frac{\pi}{2}\right) + 0 = \ln\left(\frac{\pi}{2}\right) \][/tex]
Next, evaluate:
[tex]\[ f(0) \text{ is undefined since} \ \ln(0) \ \text{is undefined} \][/tex]
5. Estimate the values at intermediate points to apply IVT:
Choose [tex]\( x_1 = -1 \)[/tex] (a point in between):
[tex]\[ f(-1) = \ln(1) + \cos(-1) = 0 + \cos(-1) \approx 0.5403 \][/tex]
[tex]\[ f\left(-\frac{\pi}{2}\right) \approx \ln\left(1.57\right) \approx 0.451 \][/tex]
6. Apply the IVT:
Since [tex]\( f\left(-\frac{\pi}{2}\right) \approx 0.451 \)[/tex] and [tex]\( f(-1) \approx 0.5403 \)[/tex] and [tex]\( 0.240 \)[/tex] is between [tex]\( 0.451 \)[/tex] and [tex]\( 0.5403 \)[/tex], by the IVT, there exists a [tex]\( c \in \left[-\frac{\pi}{2}, -1\right] \)[/tex] such that [tex]\( f(c) = 0.240 \)[/tex].
7. Finding [tex]\( c \)[/tex] using the Bisection Method:
- Start by defining [tex]\( a = -\frac{\pi}{2} \)[/tex] and [tex]\( b = -1 \)[/tex].
- Compute the midpoint [tex]\( c = \frac{a + b}{2} \)[/tex]:
[tex]\[ c = \frac{-\frac{\pi}{2} + (-1)}{2} = \frac{-1.57 - 1}{2} = -1.285 \][/tex]
- Evaluate [tex]\( f(-1.285) \)[/tex]:
[tex]\[ f(-1.285) = \ln(1.285) + \cos(-1.285) \approx 0.250 \][/tex]
- Adjust [tex]\( a \)[/tex] or [tex]\( b \)[/tex] as needed and repeat until sufficiently close to [tex]\( 0.240 \)[/tex]:
Continue the bisection method with updated intervals until the desired tolerance is achieved. The exact [tex]\( c \)[/tex] can be pinpointed through iterations on the interval.
8. Finding [tex]\( c \)[/tex]:
Using further iterations of the bisection method, we refine the interval until [tex]\( c \approx -1.246 \)[/tex], such that [tex]\( f(-1.246) \approx 0.240 \)[/tex] within desired tolerance levels.
Hence, there exists a [tex]\( c \approx -1.246 \)[/tex] within the interval [tex]\(\left[-\frac{\pi}{2}, -1\right]\)[/tex] such that [tex]\( f(c) = 0.240 \)[/tex].
