Question

Derivatives from limits The following limits represent \(f^{\prime}(a)\) for some function \(f\) and some real number \(a\) a. Find a possible function \(f\) and number \(a\). b. Evaluate the limit by computing \(f^{\prime}(a)\). $$\lim _{h \rightarrow 0} \frac{\sqrt{9+h}-\sqrt{9}}{h}$$

Short answer

Question: Identify a function f and a real number a such that the given limit represents the derivative of the function at a: $$\lim_{h \rightarrow 0} \frac{\sqrt{9+h}-\sqrt{9}}{h}$$ Answer: A possible function \(f(x) = \sqrt{9}\) and number \(a = 0\). The limit evaluating the derivative is 0, so the answer is: $$\lim_{h \rightarrow 0} \frac{\sqrt{9+h}-\sqrt{9}}{h} = 0$$

Step-by-step solution

Find the function \(f(x)\)

We know that \(f(a + h) = \sqrt{9 + h}\) and \(f(a) = \sqrt{9}\). It seems reasonable to guess that our function is of the form \(f(x) = \sqrt{x} + C\) for some constant \(C\). The \(f(a+h)\) suggests that the inside of the square root should be \(9 + h\), which we can rewrite as \((a^2 + 9 - a^2) + h\). Thus, our function is likely: $$f(x) = \sqrt{x^2 + 9 - x^2}$$ which simplifies to: $$f(x) = \sqrt{9}$$

Find the number \(a\)

Since we have found the function \(f(x)\), we can now find the number \(a\). As per the given information, the limit represents the derivative \(f'(a)\), and we know that \(f(a) = \sqrt{9} = 3\). Since in our function \(f(x) = \sqrt{9}\), this is true for all values of \(a\). So, \(a\) can be any real number. For simplicity, we can choose a = 0.

Find the derivative \(f'(a)\)

Now that we have our function \(f(x) = \sqrt{9}\) and our value \(a = 0\), we can differentiate the function to find the derivative: $$f'(x) = 0$$ Since the function is a constant and the derivative of a constant is zero.

Evaluate the limit by computing \(f'(a)\)

The final step is to evaluate the given limit using the derivative we found in step 3. Since \(f'(a) = 0\), the limit representing \(f'(a)\) is: $$\lim _{h \rightarrow 0} \frac{\sqrt{9+h}-\sqrt{9}}{h} = 0$$ In conclusion: a. A possible function \(f(x) = \sqrt{9}\) and number \(a = 0\). b. The limit evaluating the derivative is 0, so the answer is: $$\lim _{h \rightarrow 0} \frac{\sqrt{9+h}-\sqrt{9}}{h} = 0$$

Key concepts

calculus

Calculus is a branch of mathematics that deals with rates of change (differential calculus) and the accumulation of quantities (integral calculus). At its core, calculus helps us understand the behavior of functions by breaking them down into infinitely small pieces.

This concept was developed independently by Sir Isaac Newton and Gottfried Wilhelm Leibniz in the 17th century. Since its inception, calculus has found applications in science, engineering, economics, statistics, and many other fields where a quantitative analysis of change is required.

Two fundamental principles, the derivative and the integral, are the cornerstones of calculus. The derivative measures how a function changes as its input changes, while the integral sums the values of a function across a range to find the whole.

limit definition of derivative

The limit definition of derivative is a formal way of stating the derivative of a function at a particular point. It is based on the idea of finding the slope of the tangent line to the function at that point. As such, it is fundamental to the concept of differentiation.

The derivative of a function at a point \( a \), denoted as \( f'(a) \), is defined by the limit: \[ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \]
This expression represents the rate of change of the function \( f(x) \) as \( x \) approaches \( a \). The \( h \) in the formula represents a small change in \( x \), and as it gets smaller and smaller, the quotient approaches the function's instantaneous rate of change at \( a \).

differentiation

Differentiation is the process of finding the derivative of a function, which is the measure of how the output of the function changes as its input changes. In simpler terms, it tells us the rate at which one quantity changes with respect to another.

Take the function \( f(x) = x^2 \), for example. To differentiate this function, we apply rules such as the power rule, which in this case gives us a derivative \( f'(x) = 2x \). This result tells us that for every unit increase in \( x \), the value of \( f(x) \) increases by \( 2x \) units.

In the context of the exercise provided, differentiation allows us to find the slope of the function at any given point, which is crucial to solving problems involving rates of change in various physical and abstract systems.

evaluating limits

Evaluating limits is a foundational skill in calculus, particularly when working with derivatives. Limits help us understand the behavior of functions as inputs approach a certain value, including at points where functions may not be explicitly defined.

To evaluate a limit, we determine what value a function approaches as its input gets arbitrarily close to some number. In cases where direct substitution results in an indeterminate form, special techniques, such as factoring, expanding, or rationalizing the numerator and/or denominator may be used.

In the exercise at hand, the evaluation of the limit is used to find the derivative of the function \( f(x) = \sqrt{9} \) at \( a = 0 \). With these techniques, we can handle a wide variety of functions and understand their properties at specific points or over intervals. Limit evaluation is the gateway to understanding continuous and instantaneous rates of change in calculus.