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Use a quick approximation to estimate the derivative of the given function at the indicated point. HINT [See Example 2(a).] $$ f(x)=\frac{x^{2}}{2}+\frac{x}{4} ; x=2 $$

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: The approximate derivative of the function \(f(x) = \frac{x^2}{2} + \frac{x}{4}\) at x=2 is approximately equal to 2.75.

Step by step solution

01

Choose an increment for x value

For a quick approximation, we can choose a small increment in the x-value. Let's choose an increment of 1, i.e., Δx = 1.
02

Calculate the function values at x and x+Δx

Now, we can calculate the function values at x=2 and x=(2+Δx). For x=2, the function value is: f(2) = \(\frac{(2)^2}{2} + \frac{2}{4}\) = \(2 + 0.5\) = \(2.5\) For x=3 (since we have chosen Δx = 1), the function value is: f(3) = \(\frac{(3)^2}{2} + \frac{3}{4}\) = \(4.5 + 0.75\) = \(5.25\)
03

Calculate the ratio of differences

Now we'll find the ratio of the difference in y-values to the increment in x-values as an approximation to the derivative. Approximate Derivative ≈ \(\frac{Δf}{Δx}\) = \(\frac{f(3) - f(2)}{3 - 2}\) Approximate Derivative ≈ \(\frac{5.25 - 2.5}{1}\) = \(2.75\)
04

Result

Thus, the quick approximate derivative of the function \(f(x) = \frac{x^2}{2} + \frac{x}{4}\) at x=2 is approximately equal to 2.75.

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Key Concepts

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

Approximation methods
In mathematics, approximation methods are techniques used to find numerical solutions or estimates when an exact answer is not feasible. Especially in calculus, such methods can simplify complex computations. They are crucial for solving real-world problems where precise answers are challenging to obtain due to constraints like time or computational resources.
One common approach to approximation is by using derivatives, as they provide insights into how functions change. Calculating an exact derivative can be cumbersome or computationally intense. This is where quick approximations come in handy.
In our example with the function \(f(x) = \frac{x^2}{2} + \frac{x}{4}\) evaluated at \(x = 2\), instead of computing the exact derivative analytically, we use an approximation method by taking a small increment in \(x\). This means that we're estimating how \(f\) changes as \(x\) changes slightly, giving us a practical way to approach derivative estimation without involving more complicated calculus techniques.
Difference quotient
The **difference quotient** is a fundamental concept in calculus that provides a way to measure how much a function's value changes as its input changes. It's a stepping stone to the derivative, which is the 'instantaneous' rate of change.
The formula for the difference quotient is \( \frac{f(x+\Delta{x}) - f(x)}{\Delta{x}} \). Here, \(\Delta{x}\) is a small increment in \(x\), and \(f(x+\Delta{x})\) and \(f(x)\) are the values of the function at those points.
In our exercise, we chose \(\Delta{x} = 1\) to evaluate the difference quotient between \(x = 2\) and \(x = 3\). This gives us the approximate change in the function divided by the change in \(x\), which is \(\frac{5.25 - 2.5}{1} = 2.75\). This result is our estimated derivative, showing how the function behaves over that small interval.
Using the difference quotient allows you to understand a function's behavior in a localized region, offering a bridge to more intricate differential calculus concepts.
Calculus concepts
Calculus is the branch of mathematics dealing with continuous change. Its two main components are differentiation and integration. Differentiation focuses on finding rates of change, such as velocities or slopes, which is vital for understanding motion and optimization.
The derivative, which is a cornerstone of differentiation, tells us how a function changes at any point. It's like pinpointing the slope of a line that's just touching a curve at a specific point.
In our example, the function \(f(x) = \frac{x^2}{2} + \frac{x}{4}\) represents a curve. By calculating its derivative at \(x = 2\), we gain insight into the slope, or rate of change, at that specific point. Even though we used an approximation to find this derivative, the concept remains rooted in understanding how functions behave and vary.
In essence, calculus provides the tools necessary to describe and predict patterns of change, forming the foundation for much of modern science and engineering.

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Most popular questions from this chapter

Compute the derivative function \(f^{\prime}(x)\) algebraically. (Notice that the functions are the same as those in Exercises \(1-14 .)\) HINT [See Examples 2 and \(3 .]\) $$ f(x)=\frac{-1}{x} $$

The cost, in millions of dollars, of a 30 -second television ad during the Super Bowl in the years 1990 to 2007 can be approximated by the following piecewise linear function \((t=0\) represents 1990\():^{64}\) $$ C(t)=\left\\{\begin{array}{ll} 0.08 t+0.6 & \text { if } 0 \leq t<8 \\ 0.13 t+0.20 & \text { if } 8 \leq t \leq 17 \end{array}\right. $$ a. Is \(C\) a continuous function of \(t\) ? Why? HINT [See Example 4 of Section 10.3.] b. Is \(C\) a differentiable function of \(t ?\) Compute \(\lim _{t \rightarrow 8^{-}} C^{\prime}(t)\) and \(\lim _{t \rightarrow 8^{+}} C^{\prime}(t)\) and interpret the results. HINT [See Before we go on... after Example 5.]

Suppose the demand for a new brand of sneakers is given by $$ q=\frac{5,000,000}{p} $$ where \(p\) is the price per pair of sneakers, in dollars, and \(q\) is the number of pairs of sneakers that can be sold at price \(p\). Find \(q(100)\) and estimate \(q^{\prime}(100)\). Interpret your answers. HINT [See Example 1.]

Estimate the given quantity. \(f(x)=\ln x ;\) estimate \(f^{\prime}(1)\)

(a) use any method to estimate the slope of the tangent to the graph of the given function at the point with the given \(x\) -coordinate and \((\boldsymbol{b})\) find an equation of the tangent line in part (a). In each case, sketch the curve together with the appropriate tangent line. HINT [See Example \(2(\mathrm{~b}) .]\) $$ f(x)=2 x+4 ; x=-1 $$

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