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Chapter 1: Problem 37

Let \(f(t)=3 t+5\) a. Evaluate \(f(0)\) b. Solve \(f(t)=0\)

Short Answer

Expert verified
a. \(f(0) = 5\) b. \(t = -\frac{5}{3}\)

Step by step solution

01

Evaluate f(0) by substituting

To find \(f(0)\), substitute \(t = 0\) into the function \(f(t) = 3t + 5\).\[ f(0) = 3(0) + 5 \] Calculate \(3(0) + 5\) to find the value of \(f(0)\).
02

Perform the calculation for f(0)

Simplify the expression by calculating the multiplication and addition:\[ f(0) = 0 + 5 = 5 \] Thus, \(f(0) = 5\).
03

Set the function equal to zero

To solve for \(t\) when \(f(t) = 0\), set the function equal to zero:\[ 3t + 5 = 0 \] Solve this equation for \(t\).
04

Isolate the term with t

Subtract 5 from both sides of the equation to isolate the term with the variable \(t\):\[ 3t = -5 \]
05

Solve for t by dividing

Divide both sides of the equation by 3 to solve for \(t\):\[ t = \frac{-5}{3} \] Thus, \(t = -\frac{5}{3}\).

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Key Concepts

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

Evaluating Functions
Evaluating a function is like answering the question: "What outputs do we get for certain inputs?" When dealing with functions, every value you plug into the function is like running a test to see what comes out on the other side. In the context of our example, we start with the function \( f(t)=3t+5 \). Here is how we evaluate:
  • The function shows us what to do with any value we plug in for \( t \).
  • When the exercise asks to find \( f(0) \), it wants us to substitute 0 in the place of \( t \).
  • So, we replace \( t \) with 0: \( f(0) = 3(0) + 5 \).
By evaluating, we simplify the expression step-by-step. This means multiplying out 3 by 0, which gives 0, then adding 5, leading to \( f(0) = 5 \). This process of substituting and simplifying is crucial to understand, as it is a common task in many math problems involving functions.
Solving Linear Equations
Solving linear equations is all about finding the value of the variable that makes the equation true. Let's consider equation solving as a puzzle where the goal is to figure out which number, when used, balances the equation.For the equation \( f(t) = 3t + 5 = 0 \):
  • Step one is to set the equation equal to zero to find the value of \( t \) that makes \( f(t) \) equal zero.
  • We rearrange the equation by performing arithmetic operations: start by subtracting 5 from both sides to get \( 3t = -5 \).
  • The last step is isolating \( t \). Divide every part by 3, so \( t = \frac{-5}{3} \).
Linear equations like these are called 'linear' because they create straight lines when plotted graphically. Solving them helps us determine points of intersection or specific output values in different scenarios.
Function Notation
Function notation might sound fancy, but it's just a different way of saying "calculate something." In mathematics, a function is a relation that uniquely assigns an output to each input. In our example, 'function notation' refers to how we represent functions, in this case, \( f(t) = 3t + 5 \). Here's what each part means:
  • \( f \) is the function's name. It could be any letter like \( g \), \( h \), etc.
  • \( t \) is the variable or input of the function, typically representing a number.
  • 3t + 5 is the formula, shows the relationship of \( f \) that gives the output for each \( t \).
The function notation \( f(t) \) is also a signal of what variable is being used. This specific style helps us keep track of equations, especially when solving or evaluating multiple expressions at once. Understanding this is integral in mastering algebra and calculus.

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Most popular questions from this chapter

Select all of the following tables which represent \(y\) as a function of \(x\) and are one-to-one. a. $$ \begin{array}{|l|l|l|l|} \hline x & 3 & 8 & 12 \\ \hline y & 4 & 7 & 7 \\ \hline \end{array} $$ b. $$ \begin{array}{|l|l|l|l|} \hline x & 3 & 8 & 12 \\ \hline y & 4 & 7 & 13 \\ \hline \end{array} $$ c. $$ \begin{array}{|l|l|l|l|} \hline x & 3 & 8 & 12 \\ \hline y & 4 & 7 & 7 \\ \hline \end{array} $$

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Determine the interval(s) on which the function is increasing and decreasing. $$ k(x)=-3 \sqrt{x}-1 $$

Let \(g(p)=6-2 p\) a. Evaluate \(g(0)\) b. Solve \(g(p)=0\)
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