Can you check my answer?

Solve the separable differential equation: dy/dx=(sqrt(x))/2y
y=(2/3)x^(3/4)

Let f be the function given by f(x)=x^3-5x^2+3x+k is a constant.
a) On what intervals is f is increasing? (-oo,1/3), (3,oo)
b) On what intervals is the graph of f concave downward? (-oo,10/9)
c)Find the value of k for which f has 11 as its relative maximum.
I am not sure on this one. Here is what I think I would start this one:
11=x^3-5x^2+3x+k What would I plug in for x?

To find the value of k for which f has 11 as its relative maximum, we can begin by setting the derivative of f(x) equal to zero. This is because at relative extrema, the derivative will be zero.

So, we have f'(x) = 3x^2 - 10x + 3. Setting it equal to zero, we get:

3x^2 - 10x + 3 = 0.

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a,

where a = 3, b = -10, and c = 3.

Plugging in the values, we have:

x = (-(-10) ± sqrt((-10)^2 - 4 * 3 * 3)) / (2 * 3)
= (10 ± sqrt(100 - 36)) / 6
= (10 ± sqrt(64)) / 6
= (10 ± 8) / 6.

This gives us two possible values for x: x = 3/3 = 1 or x = 2.

Now, to determine which of these values of x corresponds to a relative maximum, we can investigate the concavity of f(x). Since (b) asks for the intervals where the graph of f is concave downward, we need to find the intervals where the second derivative of f(x) is negative.

To do this, we need to find f''(x). Taking the derivative of f'(x) = 3x^2 - 10x + 3, we get:

f''(x) = 6x - 10.

Now, we set f''(x) < 0 to identify the intervals where f is concave downward:

6x - 10 < 0.

Solving this inequality, we have:

6x < 10,
x < 10/6,
x < 5/3.

Therefore, the graph of f(x) is concave downward for x < 5/3.

To determine which value of x, 1 or 2, satisfies this condition, we can plug them into the inequality:

For x = 1: 1 < 5/3 (true),
For x = 2: 2 < 5/3 (false).

Hence, the graph of f(x) is concave downward on the interval (-∞, 5/3).

Finally, to find the value of k for which f has 11 as its relative maximum, we substitute x = 1 into the original equation:

f(1) = (1)^3 - 5(1)^2 + 3(1) + k
= 1 - 5 + 3 + k
= -1 + k.

Since we want f(1) = 11, we have:

-1 + k = 11.

Solving for k, we find:

k = 11 + 1
= 12.

Therefore, the value of k for which f has 11 as its relative maximum is 12.