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Chapter 3: Problem 26

Evaluate the given functions. The values of the independent variable are approximate. $$\text { Given } g(t)=\sqrt{t+1.0604}-6 t^{3}, \text { find } g(0.9261)$$.

Short Answer

Expert verified
\( g(0.9261) \approx -3.3551 \).

Step by step solution

01

Substitute the Value

Substitute the value of \( t \) into the function \( g(t) \). We are finding \( g(0.9261) \), so substitute \( t = 0.9261 \) into \( g(t) = \sqrt{t+1.0604} - 6t^3 \).
02

Evaluate the Square Root Term

Calculate \( \sqrt{0.9261 + 1.0604} = \sqrt{1.9865} \). Using a calculator, find that \( \sqrt{1.9865} \approx 1.4095 \).
03

Evaluate the Cubic Term

Calculate the cubic term: \( 6 \times (0.9261)^3 \). Raise \( 0.9261 \) to the power of 3 to get \( (0.9261)^3 = 0.7941 \). Then, multiply by 6 to get \( 6 \times 0.7941 = 4.7646 \).
04

Combine the Results

Now, substitute the evaluated terms back into the equation: \( 1.4095 - 4.7646 \). Simplify this to get \( 1.4095 - 4.7646 = -3.3551 \).

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Key Concepts

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

Square Root Calculation
Calculating the square root is a fundamental mathematical operation. It involves finding a number which, when multiplied by itself, results in the original number. The square root of a positive number always has two possible values: one positive and one negative. However, by convention, the principal square root is considered positive.

In this exercise, we encountered the square root calculation when we evaluated the term \( \sqrt{0.9261 + 1.0604} = \sqrt{1.9865} \). This computation can be performed using a calculator for precise results. A useful tip when performing calculations such as this is to first ensure the sum inside the square root is correct.
  • Check the values you are adding.
  • Use parentheses to clarify calculations.
  • Recheck calculator results if needed.
Cubic Polynomial
A cubic polynomial term typically looks like \( at^3 \) where a is a coefficient, which here is 6. These polynomials involve terms to the power of three, representing a curved graph when plotted. The characteristics of cubic polynomials make them unique as they can take on a variety of shapes.

In this problem, we calculate the cubic polynomial term \( 6 \times (0.9261)^3 \). Here is how you can approach this:
  • First, compute \( (0.9261)^3 \), which means multiplying the number by itself twice: \((0.9261) \times (0.9261) \times (0.9261)\).
  • The result is then multiplied by 6, following the given term \(6t^3\).
Understanding cubic terms allows you to manipulate and evaluate expressions that involve these operations.
Substitution Technique
The substitution technique is a straightforward yet powerful method for solving mathematical problems, particularly those involving functions. This approach involves replacing a variable in an expression or equation with a given value, which simplifies the computation process.

In our given exercise, we substituted \( t = 0.9261 \) into the function \( g(t) = \sqrt{t+1.0604} - 6t^3 \). Here is why substitution is useful:
  • It reduces a complex expression to one that is easier to evaluate.
  • It helps directly compute the value for specific instances of variables.
Substitution is often used in algebra to simplify expressions at various steps during problem-solving, making it a valuable tool in a student's mathematical toolkit.

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Most popular questions from this chapter

Graph the indicated functions. The resistance \(R\) (in \(\Omega\) ) of a resistor as a function of the temperature \(T\) (in \(^{\circ} \mathrm{C}\) ) is given by \(R=250(1+0.0032 T) .\) Plot \(R\) as a function of \(T\)

Graph the indicated functions. A measure of the light beam that can be passed through an optic fiber is its numerical aperture \(N\). For a particular optic fiber, \(N\) is a function of the index of refraction \(n\) of the glass in the fiber, given by \(N=\sqrt{n^{2}-1.69} .\) Plot the graph for \(n \leq 2.00\)

Use the following table that gives the rate R of discharge from a tank of water as a function of the height \(H\) of water in the tank. For Exercises 19 and \(20,\) plot the graph and find the values from the graph. For Exercises 21 and \(22,\) find the indicated values by linear interpolation. $$\begin{array}{l|c|c|c|c|c|c|c}\text {Height}(\mathrm{ft}) & 0 & 1.0 & 2.0 & 4.0 & 6.0 & 8.0 & 12 \\ \hline \text {Rate }\left(\mathrm{ft}^{3} / \mathrm{s}\right) & 0 & 10 & 15 & 22 & 27 & 31 & 35 \end{array}$$. (a) For \(R=34 \mathrm{ft}^{3} / \mathrm{s},\) find \(H\). (b) For \(H=6.4 \mathrm{ft}\), find \(R\).

Solve the given problems. The electric power \(P\) (in W) dissipated in a resistor of resistance \(R\) (in \(\Omega\) ) is given by the function \(P=\frac{200 R}{(100+R)^{2}}\). Because \(P=f(R),\) find \(f(R+10)\) and \(f(10 R)\).

Solve the given problems. The concentration \(C\) (in \(\mathrm{mg} / \mathrm{L}\) ) of a drug in a patient's bloodstream \(t\) hours after taking a pill is given by \(C=\frac{10.0 t}{t^{2}+1.00} \cdot\) The patient should receive a second dose after the concentration has dropped below 1.50 mg/L. How long will this take?
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