Problem 28
Question
Use the given function \(f\) to find \(f(0)\) and solve \(f(x)=0\) $$f(x)=3-\frac{2}{5} x$$
Step-by-Step Solution
Verified Answer
\(f(0) = 3\) and \(x = 7.5\) for \(f(x) = 0\).
1Step 1: Substitute for f(0)
To find the value of \(f(0)\), substitute \(x = 0\) into the function \(f(x) = 3 - \frac{2}{5}x\). This gives \(f(0) = 3 - \frac{2}{5}(0) = 3\).
2Step 2: Set the Function Equal to Zero
To solve the equation \(f(x) = 0\), set the function equal to zero: \(0 = 3 - \frac{2}{5}x\).
3Step 3: Isolate x
Rearrange the equation to isolate \(x\): \(\frac{2}{5}x = 3\) by subtracting 3 from both sides and then multiplying both sides by \(\frac{5}{2}\) to solve for \(x\): \(x = \frac{5}{2} \times 3\).
4Step 4: Calculate x
Calculate the value of \(x\): \(x = \frac{5}{2} \times 3\), which simplifies to \(x = \frac{15}{2}\) or \(x = 7.5\).
Key Concepts
Function EvaluationSolving EquationsAlgebraic Manipulation
Function Evaluation
When evaluating a function, we essentially plug a given input into a function to find its corresponding output. A function, like the one given here, is an equation that relates an input value to an output value. In the exercise, we have the function \( f(x) = 3 - \frac{2}{5}x \). To evaluate \( f(0) \), we replace the variable \( x \) with 0, yielding:
\[ f(0) = 3 - \frac{2}{5}(0) = 3 \]
This tells us that the output value when the input \( x = 0 \) is the constant 3. Function evaluation is crucial because it allows us to understand how changes in the input value affect the output.
\[ f(0) = 3 - \frac{2}{5}(0) = 3 \]
This tells us that the output value when the input \( x = 0 \) is the constant 3. Function evaluation is crucial because it allows us to understand how changes in the input value affect the output.
Solving Equations
Solving an equation means finding the value of the variable that makes the equation true. In our exercise, we need to solve \( f(x) = 0 \). This means we're looking for the \( x \) value that results in the output of the function being zero.
Starting with the equation:
\[ 0 = 3 - \frac{2}{5}x \]
we aim to find the value of \( x \) that makes this true. We do this by setting the function equal to zero and solving for \( x \). This gives us a clearer picture of where the graph of the function intersects the x-axis.
Starting with the equation:
\[ 0 = 3 - \frac{2}{5}x \]
we aim to find the value of \( x \) that makes this true. We do this by setting the function equal to zero and solving for \( x \). This gives us a clearer picture of where the graph of the function intersects the x-axis.
Algebraic Manipulation
Algebraic manipulation involves rearranging equations and expressions to isolate variables or simplify expressions. In the solution to the given problem, we performed several algebraic manipulations to solve for \( x \).
Initially, we have:
\[ \frac{2}{5}x = 3 \]
By multiplying both sides by \( \frac{5}{2} \), we cancel out the fraction on the left, leaving:
\[ x = \frac{5}{2} \times 3 \]
This simplifies to:
\[ x = \frac{15}{2} \] or \( x = 7.5 \).
This step of manipulating the equation is essential when solving formulas and helps us understand how different components of an equation relate to each other.
Initially, we have:
\[ \frac{2}{5}x = 3 \]
By multiplying both sides by \( \frac{5}{2} \), we cancel out the fraction on the left, leaving:
\[ x = \frac{5}{2} \times 3 \]
This simplifies to:
\[ x = \frac{15}{2} \] or \( x = 7.5 \).
This step of manipulating the equation is essential when solving formulas and helps us understand how different components of an equation relate to each other.
Other exercises in this chapter
Problem 28
In Exercises \(21-41,\) determine analytically if the following functions are even, odd or neither. $$ f(x)=-x^{5}+2 x^{3}-x $$
View solution Problem 28
In Exercises \(21-45,\) find and simplify the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the given function. $$ f(x)=x^{3}+1 $$
View solution Problem 28
Find the distance \(d\) between the points and the midpoint \(M\) of the line segment which connects them. $$ (2 \sqrt{45}, \sqrt{12}),(\sqrt{20}, \sqrt{27}) $$
View solution Problem 29
In Exercises \(21-41,\) determine analytically if the following functions are even, odd or neither. $$ f(x)=x^{6}-x^{4}+x^{2}+9 $$
View solution