What is the minimum value of the function g(x)= 1/x * sin (pi *x) on the interval [1,2]?
I got -2/3. Is this right? thanks.
Is the function
f(x)=1/(x sin(πx))
or
f(x)=(1/x) sin(πx) [ as written ]?
On a closer look, the first equation f(x)=1/(x sin(πx)) is undefined at x=1 and x=2, so probably not what you want.
For equation 2,
f(x)=(1/x) sin(πx)
you would find
f'(x)= (%pi*cos(πx))/x - sin(πx)/x^2
equate f'(x) to zero to find the minimum, and solve for a root between x=1 and 2.
It seems that numerical method (such as Newton's method or bisection method) will be required to find the root, which is situated at x=1.43. Find f"(x) to verify that f"(1.43)>0 for a minimum. Else, you could do the check using a graphing calculator.
Note:
-3/2 is not in [1,2], so it cannot be the correction solution.
To find the minimum value of the function g(x) = (1/x) * sin(πx) on the interval [1,2], you can follow these steps:
1. Begin by finding the critical points of the function within the given interval. These occur where the derivative of the function is equal to zero.
Let's first find the derivative of g(x). Using the product rule, we have:
g'(x) = (1/x)' * sin(πx) + 1/x * (sin(πx))'
Differentiating (1/x) gives: (1/x)' = -1/x^2
The derivative of sin(πx) with respect to x is: (sin(πx))' = π * cos(πx)
Combining the derivatives, we get:
g'(x) = (-1/x^2) * sin(πx) + (1/x) * π * cos(πx)
2. Set g'(x) equal to zero and solve for x to find the critical points within the interval [1,2].
(-1/x^2) * sin(πx) + (1/x) * π * cos(πx) = 0
3. Since sin(πx) is zero at x = 0, the term involving sin(πx) becomes zero when x = 0 is within the interval [1,2]. However, x = 0 is not included in the given interval, so it is not a critical point.
4. Next, we need to check the endpoints of the interval: x = 1 and x = 2.
For x = 1:
g'(1) = (-1/1^2) * sin(π*1) + (1/1) * π * cos(π*1)
= -sin(π) + π * cos(π)
= -π
For x = 2:
g'(2) = (-1/2^2) * sin(π*2) + (1/2) * π * cos(π*2)
= -sin(2π) + (1/2) * π * cos(2π)
= 0
5. We have found the critical points x = 0 (not within the interval) and x = 2. To determine which of these points gives the minimum value, we need to evaluate g(x) at these points.
For x = 0: g(0) is not defined since 1/0 is undefined.
For x = 2:
g(2) = (1/2) * sin(π*2)
= (1/2) * sin(2π)
= 0
6. Finally, compare the values of g(x) at the critical points and the endpoints.
g(1) = (1/1) * sin(π*1)
= sin(π)
= 0
g(2) = 0 (as found before)
Comparing these values, we see that the minimum value of g(x) within the interval [1,2] is 0, which is greater than -2/3. Therefore, -2/3 is not the correct minimum value for the function g(x).