Problem 2
Question
Find the derivative at the indicated point from the graph of \(y=f(x)\). \(f(x)=-3 x ; x=-2\)
Step-by-Step Solution
Verified Answer
The derivative at \(x = -2\) is \(-3\).
1Step 1: Identify the Function
First, we identify the given function that we will be working with. The function is \(f(x) = -3x\). This is a linear function.
2Step 2: Recall the Derivative of a Linear Function
The derivative of a function \(f(x) = mx + b\), where \(m\) and \(b\) are constants, is simply the slope \(m\) of the line. Since \(f(x) = -3x\), the derivative \(f'(x)\) is \(-3\).
3Step 3: Evaluate the Derivative at the Point
Now that we know the derivative of the function is \(-3\), we evaluate it at the point \(x = -2\). For linear functions, the derivative is constant, so \(f'(x) = -3\) for any value of \(x\). Therefore, at \(x = -2\), \(f'(-2) = -3\).
Key Concepts
Linear FunctionSlopeEvaluate the Derivative
Linear Function
A linear function is a mathematical expression that creates a straight line when graphed on a coordinate plane. The general form of a linear function is given by \( f(x) = mx + b \), where:
Here, the line has a slope of \( -3 \) and passes through the origin, since \( b = 0 \). With no y-intercept other than zero, the line is diagonal through the origin. Linear functions with just one variable like this one are the building blocks for understanding relationships in algebra.
- \( m \) represents the slope of the line
- \( b \) represents the y-intercept—the point where the line crosses the y-axis
Here, the line has a slope of \( -3 \) and passes through the origin, since \( b = 0 \). With no y-intercept other than zero, the line is diagonal through the origin. Linear functions with just one variable like this one are the building blocks for understanding relationships in algebra.
Slope
The slope of a line is a measure of its steepness. It tells us how much the y-value of the function changes for a change in the x-value. For a linear function \( f(x) = mx + b \), the slope is represented by the coefficient \( m \).
In simple terms:
In simple terms:
- If the slope is positive, the line rises as it moves from left to right.
- If the slope is negative, the line falls as it moves from left to right.
- A larger absolute value of the slope means a steeper line.
Evaluate the Derivative
Evaluating the derivative of a function at a certain point gives the rate of change at that point. For linear functions, this process is simplified since the derivative is constant and equals the slope of the line.
In the exercise, we have the linear function \( f(x) = -3x \). The derivative, therefore, is the slope \( f'(x) = -3 \). This means the rate of change maintains a consistent value.
In the exercise, we have the linear function \( f(x) = -3x \). The derivative, therefore, is the slope \( f'(x) = -3 \). This means the rate of change maintains a consistent value.
- Since this function is linear, the derivative \( f'(x) = -3 \) holds true for any \( x \).
- Thus, evaluating the derivative at \( x = -2 \) simply confirms that the rate of change at this specific point also is \(-3\).
Other exercises in this chapter
Problem 1
In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(x)=(x-3)^{2} $$
View solution Problem 2
Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a
View solution Problem 2
Find the inverse of each function and differentiate each inverse in two ways: (i) Differentiate the inverse function directly, and (ii) use (4.14) to find the d
View solution Problem 2
Differentiate the functions with respect to the independent variable. \(f(x)=e^{-2 x}\)
View solution