Question

Show that for every \(a>0\) the linear approximation \(L(x)\) to the function \(f(x)=\sqrt{x}\) at \(a\) satisfies \(f(x) \leq L(x)\) for all \(x>0\).

Short answer

For any \\(a > 0\\), \\(\sqrt{x} \leq \sqrt{a} + \frac{x - a}{2\sqrt{a}}\\) for all \\(x > 0\\).

Step-by-step solution

Define the Function and the Linear Approximation

The function given is \(f(x) = \sqrt{x}\). The linear approximation of \(f(x)\) at \(x = a\) is given by the formula for linear approximation: \(L(x) = f(a) + f'(a)(x - a)\).

Find Derivative of the Function

To form the linear approximation, we need the derivative of \(f(x) = \sqrt{x}\), which is \(f'(x) = \frac{1}{2\sqrt{x}}\).

Calculate Approximation Parameters at x = a

Now evaluate the function and its derivative at \(x = a\): \(f(a) = \sqrt{a}\) and \(f'(a) = \frac{1}{2\sqrt{a}}\).

Write the Linear Approximation

Plug these values into the linear approximation formula: \(L(x) = \sqrt{a} + \frac{1}{2\sqrt{a}}(x - a)\).

Confirm Inequality for All x > 0

To show \(f(x) \leq L(x)\) for all \(x > 0\), consider the function \(g(x) = L(x) - f(x)\). We have: \\[g(x) = \left(\sqrt{a} + \frac{1}{2\sqrt{a}}(x - a)\right) - \sqrt{x} = \sqrt{a} + \frac{x - a}{2\sqrt{a}} - \sqrt{x}\]Since \(g(x) \) is based on the linear approximation which is tangent to \(f(x) \) at \(x = a \), and because the square root function is concave, the linear approximation will be an overestimate for all \(x > 0\). Therefore, \(g(x) \geq 0 \) for all \(x > 0\).

Key concepts

Derivative of a Function

The derivative of a function is a crucial concept in calculus. It represents the rate of change of the function's value with respect to a change in its input value. In simpler terms, it tells us how the function behaves when the input changes slightly.
For the square root function \(f(x) = \sqrt{x}\), the derivative is \(f'(x) = \frac{1}{2\sqrt{x}}\). This derivative helps us understand how fast the function \(f(x)\) is increasing as \(x\) increases.
To apply this knowledge, we often use the derivative in creating linear approximations. Linear approximations help us estimate a function's value close to a specific point \(x = a\). By knowing \(f'(a)\), we can construct a tangent line at \(x = a\), which can serve as a good estimate of the function near \(a\).

Inequality Proof

Proving inequalities forms a vital part of understanding how different mathematical expressions relate to each other. In the exercise, we need to prove that for all \(x > 0\), the linear approximation \(L(x)\) is greater than or equal to the function \(f(x) = \sqrt{x}\).
To confirm \(f(x) \leq L(x)\), we consider the function \(g(x) = L(x) - f(x)\). If \(g(x) \geq 0\) for all \(x > 0\), our initial inequality is proven. The quadratic function, with its gentle curve, tends to lie below its tangent line given it is concave, ensuring \(g(x)\) stays non-negative.

• Linear approximation \(L(x)\) acts as an overestimate because it's tangent to the function at \(x = a\).
• The construction of \(g(x)\) tells us whether the linear approximation exceeds or matches the function's value.

Concave Function

Concave functions have unique characteristics. They curve downwards, like an upside-down bowl, meaning their tangent lines touch the graph at only one point and lie above the actual graph (except at the point of tangency).
Understanding concavity helps us justify the inequality of \(f(x) \leq L(x)\). The square root function \(f(x) = \sqrt{x}\) is a concave function because as we plot it, the curve starts steeper and gradually flattens out. This means the slope is decreasing—hence, the function is concave.

• The second derivative test can generally confirm concavity, but for a basic understanding, direct observation of the curve's nature suffices.
• In this context, knowing \(f(x)\) is concave lets us assure that at any \(x > 0\), \(L(x)\) stays above \(f(x)\).

Square Root Function

The square root function, \(f(x) = \sqrt{x}\), is essential in many mathematical applications. Its characteristics include starting at point (0,0) with an initial steepness then gradually leveling off as \(x\) increases.
This function is always increasing for \(x > 0\) due to its positive derivative. The linear approximation of this function creates a straight line estimate that is quite useful when approximating the curve near any point \(x = a\).
In practice, this approximation helps engineers and scientists make quick estimates, especially when working with complex calculations. Its concave nature and easy differentiation make it a perfect example for educational purposes when learning about function approximation and related inequalities.