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,inf] 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.
A. To find the equation of the tangent line to f(x) at the point (1,2), we need to determine both the slope and the y-intercept of the line.
1. Find the slope: For any point (x,y) on the graph of f(x), the slope of the tangent line is given by dy/dx. In this case, dy/dx = (-6x^2)/y. Since we want the slope at the point (1,2), we need to evaluate dy/dx at x=1 and y=2.
dy/dx = (-6(1)^2)/2 = -3
2. Find the y-intercept: We already have a point on the line, (1,2). Using the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept, we can substitute the values for the slope and the coordinates of the point to solve for b.
2 = -3(1) + b
b = 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 find an equation that satisfies dy/dx = (-6x^2)/y while also meeting the given range condition.
To solve this, we can separate the variables and integrate:
dy/y = (-6x^2)dx
Integrating both sides:
ln|y| = -2x^3 + C1 (where C1 is the constant of integration)
We can eliminate the absolute value by taking the exponential of both sides:
e^(ln|y|) = e^(-2x^3 + C1)
|y| = e^C1 * e^(-2x^3)
Now, since the range of f(x) is [0,inf], we can rewrite the equation as:
y = k * e^(-2x^3) (where k is a positive constant)
To find the value of k, we can use the given initial condition f(1) = 2:
2 = k * e^(-2(1)^3)
2 = k * e^(-2)
Solving for k:
k = 2 / e^(-2)
Therefore, the function f(x) is given by:
f(x) = (2 / e^(-2)) * e^(-2x^3)
C. To determine the domain of the function, we need to find the restrictions on x where the function is defined.
In this case, since there are no restrictions on x in the given equation dy/dx = (-6x^2)/y, the domain of the function is all real numbers.