Problem 247 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If the derivative of the function \(\begin{aligned} f(x) &=a x^{2}+b, \quad x<-1 \\ &=b x^{2}+a x+4, \quad x \geq-1 \end{aligned}\) is everywhere continuous, then find \(a\) and \(b\).

Step-by-Step Solution

Answer

The values of \(a\) and \(b\) are \(4/5\) and \(6/5\) respectively.

Step 1: Write down the values of function at the given point

First, write down the values of \(f(x)\) at \(x=-1\) for both equations separately:\( f(-1) = a(-1)^2 + b = a + b \f(-1) = b(-1)^2 + a(-1) + 4 = b - a + 4\)

Step 2: Make function continuous at \(x=-1\)

For \(f(x)\) to be continuous at \(x=-1\), the values from both parts of the piecewise function must be equal at \(x=-1\). Therefore, set \(a + b = b - a + 4\). Solving for \(a\), we get \(a = 2 - b\).

Step 3: Write down the derivative of the function

Write down the derivative of \(f(x)\) for both equations separately:\( f'(x) = 2ax \quad for \quad x < -1 \f'(x) = 2bx + a \quad for \quad x \geq -1 \)

Step 4: Write down the values of derivative at the given point

Now, write down the values of \(f'(x)\) at \(x=-1\) for both equations:\( f'(-1) = 2a(-1) = -2a \ f'(-1) = 2b*(-1) + a = -2b + a \)

Step 5: Make derivative continuous at \(x=-1\)

For \(f'(x)\) to be continuous at \(x=-1\), the value of the derivative from both parts of the piecewise function must be equal at \(x=-1\). Hence, set \(-2a = -2b + a\). Solving for \(b\), we get \(b = 3a/2\).

Step 6: Solve the two equations

Now we have two equations with two unknowns \(a\) and \(b\):\( a = 2 - b \ b = 3a/2 \) Substituting \(b = 3a/2\) into \(a = 2 - b\), we get \(a = 2 - 3a/2 \). Solving for \(a\), we get \(a = 4/5\). Substituting \(a = 4/5\) into \(b = 3a/2\), we get \(b = 6/5\).

Key Concepts

Piecewise Functions

Piecewise functions are special types of functions defined by different expressions based on the input value. These functions are particularly useful in modeling situations where a rule or behavior changes at different parts of a domain. For example, a piecewise function can be used to represent a tax rate that changes based on income brackets, or even physical phenomena like temperature changes across different materials.

When dealing with piecewise functions, it's essential to understand their continuity. A function is continuous at a point if its value doesn't "jump" at that point, meaning there is no sudden change. Mathematically, this means for a piecewise function to be continuous at a boundary, the left-hand limit and the right-hand limit, along with the function's value at that point, all need to be equal.

To visualize, imagine you are drawing the function without lifting your pencil from the paper. If you manage to do this smoothly across all points, including boundary points, the function is continuous.

Derivative Continuity

The concept of derivative continuity requires that the derivative of a function does not "jump" or experience sudden changes. Just like with piecewise functions, ensuring continuity of the derivative at boundary points is crucial, particularly in physics and engineering, where the rate of change, or derivative, often corresponds to something physical, like speed or force.

To achieve derivative continuity in a piecewise function, the derivatives of each piece must match at the boundary point. If you imagine the derivative as a slope, this means ensuring a smooth transition from one slope to the other at the connecting point. This means that both the direction and steepness need to be aligned perfectly.

In practice, set the derivatives equal at the boundary to find any unknown parameters. This helps maintain that smooth transition and ensures that your function behaves predictably across its entire domain.

Differential Calculus

Differential calculus focuses on understanding how functions change. It is the study of derivatives, which quantify the rate of change of a function concerning its variables. In simple terms, derivatives can show how quickly or slowly something is changing at any given point.

The beauty of differential calculus is its ability to address myriad problems, from predicting how fast yogurt will fall off a spoon (physical modeling) to how profits might climb in response to increased sales (economics). For students, learning about derivatives involves tackling slopes of curves, rates of change, and approximate values for functions, among other things.

In the context of solving problems like the one given, differential calculus allows you to set up equations that help ensure both function and derivative are continuous. By adding and solving these equations, you can identify the unknown parameters of a function. Understanding these principles not only ensures accurate solutions but also enhances your ability to apply calculus concepts in real-world contexts.