Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that for every \(a\) the linear approximation \(L(x)\) to the function \(f(x)=x^{2}\) at \(a\) satisfies \(L(x) \leq f(x)\) for all \(x\).

Step-by-Step Solution

Answer

\( L(x) \leq f(x) \) because \((x-a)^2 \geq 0\) for all \( x \).

Step 1: Write the linear approximation formula

The linear approximation to a function \( f(x) \) at a point \( a \) is given by: \[ L(x) = f(a) + f'(a)(x-a) \] where \( f'(a) \) is the derivative of \( f(x) \) evaluated at \( a \).

Step 2: Find the derivative of \( f(x) = x^2 \)

Compute the derivative of \( f(x) = x^2 \), denoted as \( f'(x) \). Using the power rule, \( f'(x) = 2x \).

Step 3: Write the specific linear approximation \( L(x) \) for \( f(x) = x^2 \)

Substitute \( f(a) \) and \( f'(a) \) into the linear approximation formula:\[ L(x) = a^2 + 2a(x-a) \]Simplifying gives: \[ L(x) = 2ax - a^2 \]

Step 4: Prove \( L(x) \leq f(x) \) for all \( x \)

We need to show \( 2ax - a^2 \leq x^2 \). Let's rearrange:\( x^2 - 2ax + a^2 \geq 0 \).This can be recognized as \((x-a)^2 \geq 0\), which is always true since a squared term is non-negative. Therefore, \( L(x) \leq f(x) \) for all \( x \).

Step 5: Conclusion

We demonstrated that \((x-a)^2 \geq 0\) holds for all real numbers \(x\), hence the linear approximation \( L(x) \leq f(x) \) is valid for all \( x \).

Key Concepts

Derivative

The derivative of a function is a fundamental concept in calculus, which measures how a function changes as its input changes. Imagine it as the "slope" of the curve at any given point, indicating the rate of change. For any function \( f(x) \), the derivative, often denoted as \( f'(x) \), can be thought of as how much \( f(x) \) increases or decreases with a small increase in \( x \).
To find the derivative of a function like \( f(x) = x^2 \), we use the power rule. This rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Applying this, the derivative of \( x^2 \) becomes \( 2x \).
This derivative \( f'(x) = 2x \) is key in constructing the linear approximation at a specific point, \( a \), as it provides the slope necessary for the tangent line approximation.

Quadratic Function

A quadratic function is a type of polynomial function where the highest degree is two, usually expressed as \( f(x) = ax^2 + bx + c \). These functions produce parabolic shapes when graphed, opening either upwards or downwards depending on the sign of the coefficient \( a \).
For the function \( f(x) = x^2 \), it is a simple quadratic function with its vertex at the origin (0,0). Since the coefficient of \( x^2 \) is positive, the parabola opens upwards, indicating that it has a minimum point at the vertex.
The nature of quadratic functions is pivotal when comparing linear approximations to the original function. Linear approximations essentially attempt to "flatten" the curve locally at a point \( a \), simplifying the function to understand its behavior near that point.

Inequalities

Inequalities are mathematical expressions used to compare values, showing whether one value is greater, less than, or approximately equal to another. They are useful in establishing relationships between different functions at particular points.
In the given problem, we explored the inequality \( L(x) \leq f(x) \) for all \( x \). To demonstrate this, we rewrote the expression as \( (x-a)^2 \geq 0 \). Because the square of any real number is always non-negative, this inequality holds true for all \( x \).
This result shows that the linear approximation \( L(x) \), which approximates the quadratic function \( f(x) = x^2 \) at any point \( a \), will never exceed the value of the function itself. Understanding this form of inequality helps in grasping why linear approximations can be trusted as close estimations of the function around the approximated point.