Let f be a real-valued differentiable function on R such that f(1) = 1. If the y-intercept of the ta

Q: Wait, why do we set $X=0$ to find the $y$-intercept?

That is a great question! By definition, any point on the $y$-axis has an $x$-coordinate of $0$. Since the tangent line is a straight line, its equation is $Y - y = f'(x)(X - x)$. When we set the general coordinate $X$ to $0$, we are finding the specific $Y$ value where that line crosses the vertical axis. It is the geometric definition of an intercept.

Q: Why is the differential equation $y - x \frac{dy}{dx} = x^3$ and not $y - x \frac{dy}{dx} = -x^3$?

The problem states that the $y$-intercept is equal to the cube of the abscissa, which is $x^3$. When we derive the intercept from the tangent equation $Y - y = f'(x)(X - x)$ by setting $X=0$, we get $Y = y - x f'(x)$. Equating this to $x^3$ gives $y - x f'(x) = x^3$. The sign is positive because the problem explicitly defines the intercept as $x^3$.

Q: How do I know when to use the Integrating Factor method?

Whenever you see a differential equation that can be rearranged into the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P$ and $Q$ are functions of $x$ alone, it is a linear differential equation. The Integrating Factor method is the standard, most powerful tool for solving these. If you see a $y$ and a $\frac{dy}{dx}$ term, always check if it fits this linear structure!

Q: Why do we need the condition $f(1)=1$?

When we integrate both sides of the differential equation, we introduce an arbitrary constant of integration, $C$. Without a boundary condition like $f(1)=1$, we would have an infinite family of curves. The condition $f(1)=1$ acts as a 'pin' that locks our curve to a specific point, allowing us to calculate the exact value of $C$ and find the unique function $f(x)$.

Question

Let $f$ be a real-valued differentiable function on $\mathbb{R}$ 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 $P$, find the value of $f(-3)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a curve defined by $y = f(x)$. You pick a random point $P(x, y)$ and draw a tangent line. This line is not just a random stroke; it is a geometric entity that encodes the behavior of the function at that exact moment.

The problem states that the $y$-intercept of this tangent is always $x^3$. This is a constraint that links the function's value, its slope, and its position.

To start, we write the equation of the tangent line at $P(x, y)$ using the point-slope form:
$$Y - y = f'(x)(X - x)$$
Here, $(X, Y)$ are the coordinates of any point on that tangent line.

To find the $y$-intercept, we look for the point where the line hits the $y$-axis, which happens when $X = 0$. Substituting $X = 0$ into our tangent equation, we get:
$$Y - y = f'(x)(0 - x)$$
$$Y = y - x f'(x)$$
This $Y$ is our $y$-intercept.

### The Birth of a Differential Equation

Now, we bridge the gap between geometry and calculus. The problem explicitly states that this $y$-intercept is $x^3$. So, we equate our expression to $x^3$:
$$y - x f'(x) = x^3$$

Since $f'(x)$ is just $\frac{dy}{dx}$, we have:
$$y - x \frac{dy}{dx} = x^3$$
This is a first-order linear differential equation.

Rearranging the terms, we get:
$$x \frac{dy}{dx} - y = -x^3$$
To put this into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $x$:
$$\frac{dy}{dx} - \frac{1}{x}y = -x^2$$
Now, we can clearly see that $P(x) = -\frac{1}{x}$ and $Q(x) = -x^2$.

### The Power of the Integrating Factor

We are now in the realm of standard linear differential equations. The key to solving these is the Integrating Factor ($I.F.$), defined as $e^{\int P(x) dx}$.

Substituting our $P(x)$, we get:
$$I.F. = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$$

With our $I.F.$ in hand, the general solution is given by:
$$y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$$
Plugging in our values, we get:
$$\frac{y}{x} = \int (-x^2) \cdot \frac{1}{x} dx$$
$$\frac{y}{x} = \int -x dx$$

The integration is straightforward:
$$\frac{y}{x} = -\frac{x^2}{2} + C$$

### Finding the Specific Path

We have the general solution, but we need the specific function. We use the boundary condition $f(1) = 1$, which means when $x = 1$, $y = 1$.

Substituting these into our equation:
$$1 = -\frac{1^2}{2} + C$$
$$1 = -\frac{1}{2} + C$$
Solving for $C$, we find $C = 1 + \frac{1}{2} = \frac{3}{2}$.

Now we have the complete function:
$$\frac{y}{x} = -\frac{x^2}{2} + \frac{3}{2}$$
Multiplying by $x$, we get:
$$f(x) = -\frac{x^3}{2} + \frac{3x}{2}$$

### Final Calculation

We have successfully navigated the calculus. The final step is to find $f(-3)$. Substituting $x = -3$ into our function:
$$f(-3) = -\frac{(-3)^3}{2} + \frac{3(-3)}{2}$$

Calculating this, we get:
$$f(-3) = -\frac{-27}{2} - \frac{9}{2}$$
$$f(-3) = \frac{27 - 9}{2} = \frac{18}{2} = 9$$

The final answer is **9**.

Enter Numerical Value:

Visualized Solution

Visualizing the Tangent and Intercept

Equation of the Tangent

Finding the $y$ -intercept

Formulating the Differential Equation

Rearranging to Linear Form

Identifying $P$ and $Q$

Calculating the Integrating Factor

Applying the General Solution

Integrating the Expression

Using the Boundary Condition

Solving for the Constant $C$

Defining the Function $f (x)$

Finding the Final Value $f (- 3)$

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Birth of a Differential Equation

The Power of the Integrating Factor

Finding the Specific Path

Final Calculation

Similar Questions

Let a curve $y = f (x), x \in (0, \infty)$ pass through the points $P (1, \frac{3}{2})$ and $Q (a, \frac{1}{2})$ . If the tangent at any point $R (b, f (b))$ to the given curve cuts the $y$ -axis at the point $S (0, c)$ such that $b c = 3$ , then $(P Q)^{2}$ is equal to _____.

For a differentiable function $f : R \to R$ , suppose $f^{'} (x) = 3 f (x) + α$ , where $α \in R, f (0) = 1$ and $lim_{x \to - \infty} f (x) = 7$ . Then $9 f (- lo g_{e} 3)$ is equal to_________

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Let $f$ be a differentiable function in the interval $(0, \infty)$ such that $f (1) = 1$ and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $2 f (2) + 3 f (3)$ is equal to _______

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0$ If $f (x) = 1$ , then $x$ is equal to :

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

Let $f : [1, \infty) \to R$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is

Let $y = y (x)$ be the solution of the differential equation $x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e$ . Then $y (1)$ is equal to

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Enter Numerical Value:

Visualized Solution

Visualizing the Tangent and Intercept

Equation of the Tangent

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-intercept

Formulating the Differential Equation

Rearranging to Linear Form

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculating the Integrating Factor

Applying the General Solution

Integrating the Expression

Using the Boundary Condition

Solving for the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Finding the Final Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−3)

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Birth of a Differential Equation

The Power of the Integrating Factor

Finding the Specific Path

Final Calculation

Similar Questions

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Finding the $y$ -intercept

Identifying $P$ and $Q$

Solving for the Constant $C$

Defining the Function $f (x)$

Finding the Final Value $f (- 3)$

For a differentiable function $f : R \to R$ , suppose $f^{'} (x) = 3 f (x) + α$ , where $α \in R, f (0) = 1$ and $lim_{x \to - \infty} f (x) = 7$ . Then $9 f (- lo g_{e} 3)$ is equal to_________

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function, If $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$ for all $x \geq 1$ , then the value of $f (3)$ is :

Let $f$ be a differentiable function in the interval $(0, \infty)$ such that $f (1) = 1$ and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $2 f (2) + 3 f (3)$ is equal to _______

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0$ If $f (x) = 1$ , then $x$ is equal to :

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Let $y = y (x)$ be the solution of the differential equation $x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e$ . Then $y (1)$ is equal to

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........