A function y = f(x) has a second order derivative f''(x) = 6(x - 1). If its graph passes through the

Q: Why do we need two constants of integration?

Think of integration as 'undoing' a derivative. Each time you undo a derivative, you lose information about the vertical shift of the function. Since we are integrating twice, we lose two pieces of information, hence two constants ($C_1$ and $C_2$) that must be recovered using the specific points provided.

Q: How do I know the slope of the tangent is 3?

The equation of the tangent is given as $y = 3x - 5$. This is in the slope-intercept form $y = mx + c$. By simple comparison, the coefficient of $x$ is the slope $m$. Since the tangent touches the curve at $x = 2$, the derivative $f'(2)$ must equal this slope.

Q: Can we use definite integration instead of indefinite?

You certainly could! Using definite integration from $2$ to $x$ would automatically incorporate the boundary conditions. However, finding the general form with constants $C_1$ and $C_2$ is often more intuitive for visualizing the 'family' of curves before pinning down the exact one.

Question

A function $y = f(x)$ has a second order derivative $f''(x) = 6(x - 1)$. If its graph passes through the point $(2, 1)$ and at that point the tangent to the graph is $y = 3x - 5$, then the function is

Options:

(A) (x + 1)^2

(B) (x - 1)^3

(C) (x + 1)^3

(D) (x - 1)^2

Sigma Percentile · Accepted Answer

### The Detective's Journey: Reconstructing the Curve

Welcome, future engineer. Today, we are not merely solving a calculus problem; we are engaging in a bit of mathematical archaeology. We have been given the 'DNA' of a function—its second derivative—and a few scattered clues about its location and orientation.

Our mission is to reconstruct the original function, $f(x)$, from these fragments. This is the essence of calculus: the ability to look at the rate of change of a rate of change and see the full, elegant curve hidden beneath.

### Phase 1: The Tangent's Whisper

Imagine you are standing on a roller coaster track. The track is the curve $f(x)$. At the specific point $(2, 1)$, you are given a tangent line, $y = 3x - 5$.

This line is a snapshot of your trajectory at that exact moment. It tells us two vital things. First, the curve passes through $(2, 1)$, so $f(2) = 1$.

Second, the slope of the curve at this point is the same as the slope of the tangent. Comparing $y = 3x - 5$ to the standard form $y = mx + c$, we immediately see that the slope $m = 3$. Therefore, we have our first boundary condition: $f'(2) = 3$. This is our anchor.

### Phase 2: The First Descent

We are given the second derivative, $f''(x) = 6(x - 1)$. To find the function, we must climb down the derivative ladder. We integrate. Let's set up the integral for the first derivative:

$$f'(x) = \int 6(x - 1) dx$$

I know that integration can sometimes feel like walking into a fog, but trust the process. Expanding the term, we get $6x - 6$. Integrating this, we get:

$$f'(x) = 6 \left( \frac{x^2}{2} - x \right) + C_1$$

$$f'(x) = 3x^2 - 6x + C_1$$

Here is where many students stumble: the constant $C_1$. Never forget it! It represents the 'vertical freedom' of the derivative. Now, we use our anchor from Phase 1. We know $f'(2) = 3$. Let's plug in $x = 2$:

$$3(2)^2 - 6(2) + C_1 = 3$$

$$12 - 12 + C_1 = 3$$

Look at that! The $12$ and $-12$ cancel out with such satisfying precision. We are left with $C_1 = 3$. Our first derivative is now fully revealed: $f'(x) = 3x^2 - 6x + 3$. If we factor out the $3$, we get $f'(x) = 3(x^2 - 2x + 1)$, which is the perfect square $3(x - 1)^2$. Beautiful, isn't it?

### Phase 3: The Final Descent

We are almost there. We have $f'(x) = 3(x - 1)^2$. To find $f(x)$, we integrate one last time:

$$f(x) = \int 3(x - 1)^2 dx$$

Using the power rule again, the integral of $(x - 1)^2$ is $\frac{(x - 1)^3}{3}$. Multiplying by the $3$ outside, the constants cancel out perfectly:

$$f(x) = (x - 1)^3 + C_2$$

Again, we introduce $C_2$. This is the final constant of integration. We use our last clue: the curve passes through $(2, 1)$, so $f(2) = 1$. Substituting these values:

$$1 = (2 - 1)^3 + C_2$$

$$1 = 1^3 + C_2$$

$$1 = 1 + C_2$$

Subtracting $1$ from both sides, we find $C_2 = 0$. The constant vanishes, leaving us with the elegant final function: **$f(x) = (x - 1)^3$**.

You have successfully reconstructed the curve. Take a moment to appreciate the symmetry. By respecting the boundary conditions and carefully navigating the integration, you have turned abstract derivatives into a concrete, tangible function. Keep this mindset—calculus is not just about rules; it is about uncovering the hidden structure of the world.

Animated Solution for Mathematics - Differential Equations: A function $y = f (x)$ has a second order derivative $f^{''} (x) = 6 (x - 1)$ . If its graph passes through the point $(2, 1)$ and at that point the tangent to the graph is $y = 3 x - 5$ , then the function is

Select Answer:

Visualized Solution

Visualizing the Setup

Slope of the Tangent

The Second Derivative

Integrating $f^{''} (x)$

Finding $f^{'} (x)$

Substituting for $C_{1}$

Solving for $C_{1}$

The Complete $f^{'} (x)$

Integrating to find $f (x)$

Executing the Integration

Using the Point $(2, 1)$

Solving for $C_{2}$

The Final Function

The Sigma Insight: Variable Separable Method

The Detective's Journey

Phase 1

Phase 2

Phase 3

Similar Questions

A curve $y = f (x)$ passes through (1, 1) and at $P (x, y)$ , tangent cuts the x-axis and y-axis at A and B respectively such that $B P : A P = 3 : 1$ , then

Let for some function $y = f (x)$ , $\int_{0}^{x} t f (t) d t = x^{2} f (x)$ , $x > 0$ and $f (2) = 3$ Then $f (6)$ is equal to

Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2)$ with $f (0) = 1$ . Let $F (x) = \int_{0}^{x^{2}} f (t) d t$ for $x \in [0, 2]$ . If $F^{'} (x) = f^{'} (x)$ for all $x \in (0, 2)$ , then $f (2)$ equals

Let $f$ be a differentiable function such that $f (1) = 2$ and $f^{'} (x) = f (x)$ for all $x \in R$ . If $h (x) = f (f (x))$ , then $h^{'} (1)$ is equal to :

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

If length of tangent at any point on the curve $y = f (x)$ intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

If a curve $y = f (x)$ passes through the point $(1, - 1)$ and satisfies the differential equation, $y (1 + x y) d x = x d y$ , then $f (- \frac{1}{2})$ is equal to :

Let $f$ be a differentiable function satisfying $f (x) = \frac{2}{3} \int_{0}^{3} f (\frac{λ ^{2} x}{3}) d λ, x > 0$ and $f (1) = 3$ . If $y = f (x)$ passes through the point $(α, 6)$ , then $α$ is equal to \_\_\_\_.

Let a smooth curve $y = f (x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\frac{- y}{x}$ . If the curve passes through the point $(1, 2)$ and $(8, 1)$ , then $∣ y (\frac{1}{8}) ∣$ is equal to

$f (x)$ and $g (x)$ are two differentiable functions on $[0, 2]$ such that $f^{''} (x) - g^{''} (x) = 0, f^{'} (1) = 2 g^{'} (1) = 4, f (2) = 3 g (2) = 9$ then $f (x) - g (x)$ at $x = 3/2$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Slope of the Tangent

The Second Derivative

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

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

Substituting for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

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

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