The equation dy/dx = -6x^2/y gives the slope at any point on the graph of f(x). The range of f(x) is [0, infinity] and f(1) = 2.

A. Find the equation of the tangent line to f(x) at the point (1,2).

B. Write the function f(x).

C. Determine the domain of the function f(x).

we know the slope is dy/dx, so at (1,2), the slope is -3.

The tangent line is thus

y-2 = -3(x-1)

y dy = -6x^2 dx
1/2 y^2 = -2x^3 + c
y = √2 √(c-2x^3)

you know that the domain of √ is non-negative numbers, so

c-2x^3 >= 0
2x^3 <= c
x <= ∛(c/2)

To compound off of what Steve said, because the point (1,2) is known, you can plug it into y to get C to be equal to 8. Thus, the domain of the function of would be [-∞, ∛2]. Let me know if you want to see where I got these numbers from

A. To find the equation of the tangent line to f(x) at the point (1,2), we can use the point-slope form of a linear equation.

First, let's find dy/dx at x = 1 by plugging in x = 1 into the given equation dy/dx = -6x^2/y:

dy/dx = -6(1)^2/2
dy/dx = -6/2
dy/dx = -3

So, the slope of the tangent line at x = 1 is -3.

Next, we can use the point-slope form y - y1 = m(x - x1), where (x1, y1) is the given point (1,2) and m is the slope (-3) to find the equation of the tangent line:

y - 2 = -3(x - 1)
y - 2 = -3x + 3
y = -3x + 5

Therefore, the equation of the tangent line to f(x) at the point (1,2) is y = -3x + 5.

B. To write the function f(x), we need to solve the given differential equation. Let's rearrange the equation dy/dx = -6x^2/y to separate variables:

y dy = -6x^2 dx

Next, we can integrate both sides:

∫y dy = ∫-6x^2 dx

Integrating both sides gives:

y^2/2 + C1 = -2x^3 + C2

where C1 and C2 are constants of integration.

Since f(1) = 2, we can substitute x = 1 and y = 2 into the equation to solve for C1 and C2:

(2)^2/2 + C1 = -2(1)^3 + C2
1 + C1 = -2 + C2

We can simplify this to:

C1 = C2 - 3

Now, substituting C2 - 3 for C1 in the equation:

y^2/2 + C2 - 3 = -2x^3 + C2

Simplifying gives:

y^2 = -4x^3 + 6

Therefore, the function f(x) is given by y = √(-4x^3 + 6).

C. The domain of the function f(x) is determined by the range of values for which the function is defined. In this case, since we have a square root function, the argument of the square root must be non-negative.

Setting -4x^3 + 6 ≥ 0 to find the range of x values, we have:

-4x^3 + 6 ≥ 0
-4x^3 ≥ -6
x^3 ≤ 3/2

Taking the cube root of both sides gives:

x ≤ (3/2)^(1/3)

Therefore, the domain of the function f(x) is (-∞, (3/2)^(1/3)].

To solve the problem, we'll start by finding the equation of the tangent line to f(x) at the point (1,2).

A. Find the equation of the tangent line to f(x) at the point (1,2):
To find the equation of a tangent line at a given point, we need to find the slope of the tangent line and then use the point-slope form of a linear equation.

Given: dy/dx = -6x^2/y and f(1) = 2.
We know that the slope of the tangent line is equal to dy/dx, so we can substitute x = 1 and y = 2 into the equation dy/dx = -6x^2/y to find the slope at the point (1,2).

dy/dx = -6(1)^2 / 2 = -6/2 = -3

Therefore, the slope of the tangent line at the point (1,2) is -3.

Now we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.

Using (x1, y1) = (1,2) and m = -3, we have:
y - 2 = -3(x - 1)

Expanding the equation, we get:
y - 2 = -3x + 3

Rearranging the equation to the standard form, we have:
3x + y = 5

Therefore, the equation of the tangent line to f(x) at the point (1,2) is 3x + y = 5.

B. Write the function f(x):
To find the function f(x), we need to integrate the given equation dy/dx = -6x^2/y with the initial condition f(1) = 2.

Integrating both sides of the equation, we get:
∫ dy/y = ∫ -6x^2 dx

Integrating, we have:
ln|y| = -2x^3 + C

To find the constant C, we'll use the initial condition f(1) = 2. Substituting x = 1 and y = 2 into the equation, we have:
ln|2| = -2(1)^3 + C
ln|2| = -2 + C
C = ln|2| + 2

Therefore, the equation becomes:
ln|y| = -2x^3 + ln|2| + 2

To eliminate the natural logarithm, we can exponentiate both sides:
|y| = e^(-2x^3 + ln|2| + 2)

Since the range of f(x) is [0, infinity], we can remove the absolute value sign:
y = e^(-2x^3 + ln|2| + 2)

Simplifying, we have:
y = 2e^(-2x^3 + 2)

Therefore, the function f(x) is given by f(x) = 2e^(-2x^3 + 2).

C. Determine the domain of the function f(x):
The domain of a function is the set of all possible input values for which the function is defined. In this case, since the expression -2x^3 + 2 is a polynomial, it is defined for all real numbers. Therefore, the domain of the function f(x) is (-∞, ∞).