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$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ f(x)=4 x^{2}-5 x+3 $$

Short Answer

Expert verified
The simplified expression is \(8x + 4h - 5\).

Step by step solution

01

Apply the Function to (x + h)

First, substitute \(x + h\) into the function \(f(x) = 4x^2 - 5x + 3\). This gives us \(f(x + h) = 4(x + h)^2 - 5(x + h) + 3\).
02

Expand the Expression

Now expand \( (x + h)^2 \) which is \( x^2 + 2xh + h^2 \). Substitute this into the function: \(f(x + h) = 4(x^2 + 2xh + h^2) - 5(x + h) + 3\).
03

Distribute the Constants

Distribute 4 and -5 into their respective terms: \(4(x^2 + 2xh + h^2) = 4x^2 + 8xh + 4h^2\) and \(-5(x + h) = -5x - 5h\). Simplifying, we have \(f(x + h) = 4x^2 + 8xh + 4h^2 - 5x - 5h + 3\).
04

Calculate the Difference f(x+h) - f(x)

Subtract \(f(x)\) from \(f(x+h)\): \(f(x+h) - f(x) = (4x^2 + 8xh + 4h^2 - 5x - 5h + 3) - (4x^2 - 5x + 3)\).
05

Simplify the Difference Expression

Combine like terms: \(8xh + 4h^2 - 5h\). The \(4x^2\), \(-5x\), and \(+3\) cancel out from both expressions.
06

Divide by h

Divide the simplified expression \(8xh + 4h^2 - 5h\) by \(h\): \(\frac{8xh + 4h^2 - 5h}{h} = 8x + 4h - 5\).
07

Final Simplification

Since \(h eq 0\), cancel \(h\) to get the final expression: \(8x + 4h - 5\).

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Key Concepts

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

Algebraic Simplification
Algebraic simplification involves breaking down complex expressions into simpler and more manageable forms. This is often done by expanding, combining like terms, and factoring where possible. In the given exercise, simplification starts when we substitute \(x + h\) into the function. To handle this, the function needs expanding which modifies a polynomial initially in the form \(f(x) = 4x^2 - 5x + 3\).
Expanding the expression \( (x+h)^2 \) yields \( x^2 + 2xh + h^2 \), and plugging these into the function results in a new polynomial that shows multiple terms, including terms containing \(h\) and \(xh\). This expansion is critical as it enables further simplification.
Next, distribute constants like 4 over the new terms to shape the expression appropriately. Once distributed, terms can be combined through addition or subtraction. It is crucial to observe like terms - such as identical powers of \(x\) and \(h\) - that can cancel each other out. The step involving subtraction of \(f(x)\) allows for like terms to cancel, simplifying the difference quotient to form a minimally reduced expression. Completing these steps gives clarity to the algebraic process and produces a simple expression that efficiently represents the given function after transformation.
Polynomial Functions
Polynomial functions are algebraic expressions comprising variables raised to various powers, each multiplied by coefficients. The leading type of problem where polynomial functions often come into play is in calculating change, such as the change over a distance or a time period.
In the exercise, the polynomial \(f(x)=4x^2-5x+3\) is a quadratic function. Key characteristics of polynomial functions are evident here as they smoothly model curves which can be used to describe parabolic patterns. The terms in polynomial functions are often organized by descending powers of \(x\), providing insight into the behavior of the graph of the function as it turns along its trajectory.
When we deal with polynomial functions, understanding how changes affect all the terms is vital. For example, substituting \(x + h\) into \(f(x)\) lets us examine how both constant and non-constant terms evolve as inputs change. After transformation, reaching simpler polynomial expressions through detailed steps shows how computations can reveal insights about original curves' behavior over specific intervals or conditions.
Rate of Change
The rate of change is a foundational concept in calculus, relating to how a quantity changes in relation to another. Here, the difference quotient \(\frac{f(x+h)-f(x)}{h}\) is a discrete method for finding the average rate of change of a function over a small interval \(h\).
This specific exercise involves a quadratic polynomial function, characteristically with a non-linear graph. Solving the difference quotient provides a linear approximation which simplifies the understanding of change in the function over the small gap \(h\). In practice, by evaluating the difference quotient, learners transition from understanding pure algebraic concepts to fundamental calculus ideas.
As \(h\) approaches zero, this quotient morphs into what is known as a derivative, delivering a precise instantaneous rate of change. Such insights reveal not just how quantities increment but also serve planning and prediction goals in real-world applications like physics and economics, where understanding quickly and precisely how one quantity intervenes upon another sets the base for complex systemic behaviors.

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Most popular questions from this chapter

63-64. BUSINESS: Straight-Line Depreciation Straight-line depreciation is a method for estimating the value of an asset (such as a piece of machinery) as it loses value ( \({ }^{\prime \prime}\) depreciates" \()\) through use. Given the original price of an asset, its useful lifetime, and its scrap value (its value at the end of its useful lifetime), the value of the asset after \(t\) years is given by the formula: $$ \begin{aligned} \text { Value }=(\text { Price })-&\left(\frac{(\text { Price })-(\text { Scrap value })}{(\text { Useful lifetime })}\right) \cdot t \\ & \text { for } 0 \leq t \leq(\text { Useful lifetime }) \end{aligned} $$ a. A farmer buys a harvester for $$\$ 50,000$$ and estimates its useful life to be 20 years, after which its scrap value will be $$\$ 6000$$. Use the formula above to find a formula for the value \(V\) of the harvester after \(t\) years, for \(0 \leq t \leq 20\). b. Use your formula to find the value of the harvester after 5 years. c. Graph the function found in part (a) on a graphing calculator on the window \([0,20]\) by \([0,50,000]\). [Hint: Use \(x\) instead of \(t\).]

True or False: \(\infty\) is the largest number.

Smoking and Income Based on a recent study, the probability that someone is a smoker decreases with the person's income. If someone's family income is \(x\) thousand dollars, then the probability (expressed as a percentage) that the person smokes is approximately \(y=-0.31 x+40\) (for \(10 \leq x \leq 100)\) a. Graph this line on the window \([0,100]\) by \([0,50]\). b. What is the probability that a person with a family income of $$\$ 40,000$$ is a smoker? [Hint: Since \(x\) is in thousands of dollars, what \(x\) -value corresponds to $$\$ 40,000 ?]$$ c. What is the probability that a person with a family income of $$\$ 70,000$$ is a smoker? Round your answers to the nearest percent.

BUSINESS: Research Expenditures An electronics company's research budget is \(R(p)=3 p^{0.25}\), where \(p\) is the company's profit, and the profit is predicted to be \(p(t)=55+4 t\), where \(t\) is the number of years from now. (Both \(R\) and \(p\) are in millions of dollars.) Express the research expenditure \(R\) as a function of \(t\), and evaluate the function at \(t=5\).

The intersection of an isocost line \(w L+r K=C\) and an isoquant curve \(K=a L^{b}\) (see pages 18 and 32 ) gives the amounts of labor \(L\) and capital \(K\) for fixed production and cost. Find the intersection point \((L, K)\) of each isocost and isoquant. [Hint: After substituting the second expression into the first, multiply through by \(L\) and factor.] $$ 5 L+4 K=120 \text { and } K=180 \cdot L^{-1} $$

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