/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Compute \(f(c+h)-f(c)\) at the i... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 19

Compute \(f(c+h)-f(c)\) at the indicated point. $$ f(x)=\sqrt{x} ; c=4 $$

Short Answer

Expert verified
\(f(c+h) - f(c) = \sqrt{4+h} - 2\)

Step by step solution

01

Substitute the Function

The exercise requires us to compute the difference \(f(c+h)-f(c)\). We know that \(f(x) = \sqrt{x}\). Begin by substituting \(f(c+h)\) and \(f(c)\) as \(f(x)\) indicates.So:\[f(c+h) = \sqrt{c+h}\quad\text{and}\quad f(c) = \sqrt{c}\]
02

Apply the Specific Point

Now, apply the given value \(c = 4\) into the expressions we found in Step 1.\[f(c+h) = \sqrt{4+h}\quad\text{and}\quad f(c) = \sqrt{4}\]
03

Calculate the Difference

Subtract \(f(c)\) from \(f(c+h)\) as indicated in the problem statement.\[f(c+h) - f(c) = \sqrt{4+h} - \sqrt{4}\]Since \(\sqrt{4}\) is a known value:\[\sqrt{4} = 2\]So the expression becomes:\[f(c+h) - f(c) = \sqrt{4+h} - 2\]

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Key Concepts

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

Function Difference
When working with functions, understanding the concept of a function difference is crucial. A function difference indicates how much a function's value changes between two points. In mathematical terms, this is often expressed as \( f(c+h) - f(c) \), where \( f(x) \) is the function in question.
  • The function difference helps identify trends and changes in the function's behavior over an interval.
  • In our exercise, we see this as the contrast between \( \sqrt{4+h} \) and \( \sqrt{4} \).
The objective is to analyze these changes to understand the function better, which is pivotal in calculus, especially in finding derivatives and understanding limits.
Square Root Function
The square root function, represented as \( f(x) = \sqrt{x} \), is a fundamental concept in mathematics.
  • This function involves finding a number which, when multiplied by itself, yields the original number, \( x \).
  • For instance, in our exercise, \( \sqrt{4} = 2 \), because \( 2 \times 2 = 4 \).
Square root functions are inherently non-linear, meaning they do not form a straight line when graphed. Instead, they curve, with the curve getting shallower as \( x \) increases. Understanding square root functions is key for solving problems that involve roots, radicals, and understanding growth patterns in natural phenomena.
Algebraic Manipulation
Algebraic manipulation is a technique used to simplify mathematical expressions and solve equations. It is essential when working with function differences and other calculus-related problems.
  • The goal is to transform an expression into a simpler or more workable form.
  • In our problem, algebraic manipulation helps simplify \( \sqrt{4+h} - 2 \).
Performing algebraic manipulation often involves applying arithmetic operations, factorization, and understanding mathematical properties. This skill is crucial in making complex problems much more manageable, allowing us to isolate certain variables or expressions to better analyze their components.

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Most popular questions from this chapter

Use the formal definition to find the derivative of $$ f(x)=\frac{1}{x+1} $$

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\ln (1+2 x) \text { at } a=0 $$

Suppose that the per capita growth rate of a population is \(2 \% ;\) that is, if \(N(t)\) denotes the population size at time \(t\), then $$\frac{1}{N} \frac{d N}{d t}=0.02$$ Suppose also that the population size at time \(t=2\) is equal to 50. Use a linear approximation to compute the population size at time \(t=2.1\).

Graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.) $$ f(x)=\left\\{\begin{array}{cl} x & \text { for } x \leq 0 \\ x+1 & \text { for } x>0 \end{array}\right. $$

Assume that \(N(t)\) denotes the size of a population at time \(t\) and that \(N(t)\) satisfies the differential equation $$ \frac{d N}{d t}=r N $$ where \(r\) is a constant. (a) Find the per capita growth rate. (b) Assume that \(r<0\) and that \(N(0)=20\). Is the population size at time 1 greater than 20 or less than \(20 ?\) Explain your answer.
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