Let f be a real valued differentiable function on R such that f(1) =1. If the y-intercept of the tangent at any point P(x,y) on the curve y=f(x) is equal to the cube of the abscissa of P, then the value of f(-3) is equal to

Options

A. 9
B. 10
C. `-9`
D. 10

Answer

To solve the problem, we need to find the value of \( f(-3) \) given the conditions in the problem statement. Let's go through the solution step by step.

### Step 1: Understand the problem
We are given that the y-intercept of the tangent at any point \( P(x, y) \) on the curve \( y = f(x) \) is equal to the cube of the abscissa of \( P \). This means that if we find the equation of the tangent line at point \( P \), the y-intercept must satisfy a specific condition.

### Step 2: Write the equation of the tangent line
The equation of the tangent line at point \( P(x, f(x)) \) can be expressed using the point-slope form:
\[ y - f(x) = f'(x)(x - x_1) \]
where \( f'(x) \) is the derivative of \( f(x) \) at point \( x \).

### Step 3: Find the y-intercept of the tangent line
To find the y-intercept, we set \( x = 0 \) in the tangent line equation:
\[ y - f(x) = f'(x)(0 - x) \implies y = f(x) - f'(x)x \]
Thus, the y-intercept \( Y \) is:
\[ Y = f(x) - f'(x)x \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the y-intercept \( Y \) is equal to the cube of the abscissa \( x \):
\[ f(x) - f'(x)x = x^3 \]
Rearranging gives us:
\[ f(x) - x^3 = f'(x)x \]

### Step 5: Rewrite the equation
We can rewrite this as:
\[ f'(x) = \frac{f(x) - x^3}{x} \]

### Step 6: Solve the differential equation
This is a first-order linear differential equation. We can separate variables:
\[ \frac{dy}{dx} = \frac{f(x) - x^3}{x} \]
Let \( y = f(x) \), then we have:
\[ \frac{dy}{dx} = \frac{y - x^3}{x} \]
This can be rearranged to:
\[ \frac{dy}{dx} + \frac{1}{x}y = x^2 \]

### Step 7: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int \frac{1}{x}dx} = e^{\ln|x|} = |x| \]
Since \( x \) is positive, we can just use \( x \).

### Step 8: Multiply through by the integrating factor
Multiplying the entire equation by \( x \):
\[ x \frac{dy}{dx} + y = x^3 \]

### Step 9: Integrate both sides
The left side can be simplified to:
\[ \frac{d}{dx}(xy) = x^3 \]
Integrating both sides gives:
\[ xy = \frac{x^4}{4} + C \]
Thus,
\[ y = \frac{x^3}{4} + \frac{C}{x} \]

### Step 10: Use the initial condition
We know that \( f(1) = 1 \):
\[ 1 = \frac{1^3}{4} + C \implies 1 = \frac{1}{4} + C \implies C = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 11: Write the function
Now we have:
\[ f(x) = \frac{x^3}{4} + \frac{3/4}{x} \]

### Step 12: Find \( f(-3) \)
Now we can find \( f(-3) \):
\[ f(-3) = \frac{(-3)^3}{4} + \frac{3/4}{-3} = \frac{-27}{4} - \frac{3}{12} \]
Converting \( \frac{3}{12} \) to a common denominator:
\[ \frac{3}{12} = \frac{1}{4} \]
Thus:
\[ f(-3) = \frac{-27}{4} - \frac{1}{4} = \frac{-28}{4} = -7 \]

### Final Answer
The value of \( f(-3) \) is \( -7 \).