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Chapter 3: Problem 9

Find the derivative of the following functions. $$f(x)=5$$

Short Answer

Expert verified
Answer: The derivative of the function $$f(x) = 5$$ is $$f'(x) = 0$$.

Step by step solution

01

Recognize the function

We are given the function $$f(x) = 5$$, which is a constant function (the output value does not depend on the input value, x).
02

Define the derivative

The derivative of a function represents the rate of change of the function with respect to the independent variable, x. In our case, we want to find the derivative of $$f(x) = 5$$ with respect to x. This can be denoted as $$f'(x)$$ or $$\frac{d}{dx}f(x)$$.
03

Calculate the derivative of a constant function

Since the function is constant and doesn't change with respect to x, the rate of change (derivative) should be 0. In other words, $$f'(x) = 0$$ for a constant function.
04

Final answer

Therefore, the derivative of the function $$f(x) = 5$$ is $$f'(x) = 0$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Function
A constant function is a special type of function where the output is the same, no matter the input. When we think of a graph of a constant function, imagine a flat, horizontal line. For example, the function \( f(x) = 5 \) is a constant function. Here, no matter what value you plug in for \( x \), the output is always 5.

This is significant because in terms of change, a constant function doesn't change at all. Whether \( x \) is 1, 10, or 1000, the result remains constant.
  • Output is always the same.
  • No matter the input, it leads to no change over time or space.

Knowing that a function is constant helps simplify calculations in calculus. We don't need complex operations to find its behavior, since it is predictably the same.
Rate of Change
The rate of change describes how much a function's output value changes as the input value changes. In simple terms, it's the speed at which something is happening. When we calculate the derivative of a function, we are finding its rate of change.

In calculus, if a function changes (like a slope on a hill), we can describe this change using derivatives. But for a constant function, like \( f(x) = 5 \), the rate of change is particularly simple. The function doesn't rise or fall regardless of changes in \( x \). Thus, the rate of change is zero.
  • Describes how fast or slow a change occurs.
  • For constant functions, the rate of change is always zero.

This concept is critical in understanding the behavior of functions in real-world scenarios, where change can indicate progress, decline, or stability.
Independent Variable
An independent variable represents the input or the cause that affects the change in a function. In many mathematical functions, this is usually denoted by \( x \). It is the variable we manipulate to see how the function鈥檚 output (dependent variable) responds.

In our example, \( f(x) = 5 \), \( x \) is the independent variable. While it has the liberty to take on any value, the output, \( f(x) \), remains unchanged because it is a constant function.
  • Cause of change in a function's output.
  • Can take any value without affecting the output in constant functions.

Understanding the independent variable helps in studying how different scenarios affect outputs, crucial in fields like science and economics where varying one factor can lead to numerous outcomes.

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Most popular questions from this chapter

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

Find \(\frac{d y}{d x},\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}\)

Tangency question It is easily verified that the graphs of \(y=x^{2}\) and \(y=e^{x}\) have no points of intersection (for \(x>0\) ), and the graphs of \(y=x^{3}\) and \(y=e^{x}\) have two points of intersection. It follows that for some real number \(2 0 \) ). Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

A 500-liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min} .\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank (in liters) is given by \(V(t)=500-0.5 t\). a. Graph the mass function and verify that \(M(0)=0\). b. Graph the volume function and verify that the tank is empty when \(t=1000\) min. c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) and \(\lim _{\theta \rightarrow 000^{-}} C(t) ?\) \(t \rightarrow 1\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\). e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\). f. For what times is the concentration of the solution increasing? Decreasing?

Find the following higher-order derivatives. $$\frac{d^{3}}{d x^{3}}\left(x^{2} \ln x\right)$$.
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