need help with some calculus problems please!!?

Question

dy/dx= (y-1)^2cos(pi*x)

there is a horizontal line y=c that satisfies this differential equation. find the value of c.


f is the function defined by f(x)= [k*square root of x] - ln (x) where k is a positive constant.
find the first and second derivatives of f(x)

then, for a certain value of the constant k, the graph f has a point of inflection on the x-axis. find this value of k.

those are 2 separate problems.

please show work!

Answer 1

The first problem is easy. If y=c, then dy/dx is everywhere 0, so we have that 0=(y-1)² cos (πx). cos (πx) is not everywhere 0, so we can divide by it to find (y-1)²=0, so y-1=0 and y=1. Therefore, c=1.

For the second problem, first find the first and second derivatives:

f(x) = k√x - ln x
f'(x) = k/(2√x) - 1/x
f''(x) = -k/4 x^(-3/2) + 1/x²

Now, if (x, f(x)) is a point of inflection, that means the second derivative is changing sign, which means f''(x) must be zero. If it is on the x-axis, this means f(x) must be zero. So this problem asks "if f''(x) = 0 at the same point that f(x)=0, what is the value of k?" First, we find where f''(x) = 0:

0 = -k/4 x^(-3/2) + 1/x²
k/4 x^(-3/2) = 1/x²
k/4 √x = 1
√x = 4/k
x=16/k²

Now, if f(x) is also equal to zero at this point, then we have:

0 = k√(16/k²) - ln (16/k²)
0 = 4 - ln (16/k²)
ln (16/k²) = 4
16/k² = e^4
16e^(-4) = k²
k = 4/e².

And we are done.

Answer 2

y = c
dy/dx = 0
dy/dx is only identically zero, when c = 1

For the second one.
f(x) = kx^1/2 -lnx
f' = 1/2 * kx^-1/2 - 1/x
f''(x) = -1/4 * k * x-(3/2) + 1/x^2
At the point of inflexion, f'' = 0
x^1/2 = 4/k
So the y cordinate must be zero for this to be on the x axis, so
f(4/k) = 4 - ln(4/k) = 0
k = 4 / e^4

Answer 3

To answer the first question, y=1.