Problem 16
Question
A function \(y=f(x)\) and an \(x\) -value \(x_{0}\) are given. (a) Find a formula for the slope of the tangent line to the graph of \(f\) at a general point \(x=x_{0}\). (b) Use the formula obtained in part (a) to find the slope of the tangent line for the given value of \(x_{0}\). $$f(x)=x^{2}+3 x+2 ; x_{0}=2$$
Step-by-Step Solution
Verified Answer
The slope of the tangent line at \( x = 2 \) is 7.
1Step 1: Understand the Problem
The goal is to find the slope of the tangent line to the curve defined by the function \( f(x) = x^2 + 3x + 2 \) at the specific point where \( x = 2 \). The slope of the tangent line is given by the derivative of the function evaluated at \( x_0 \).
2Step 2: Find the Formula for the Derivative
To find the slope of the tangent, we need the derivative of \( f(x) = x^2 + 3x + 2 \). Differentiate the function with respect to \( x \): \[ f'(x) = \frac{d}{dx}(x^2 + 3x + 2) = 2x + 3. \] This is the formula for the slope of the tangent line at any point \( x \).
3Step 3: Substitute the Given Value
Now, substitute \( x_0 = 2 \) into the derivative to find the slope of the tangent line at this point:\[ f'(2) = 2(2) + 3 = 4 + 3 = 7. \]
4Step 4: Conclude the Result
The slope of the tangent line to the graph \( f(x) = x^2 + 3x + 2 \) at the point where \( x = 2 \) is \( 7 \).
Key Concepts
Tangent LineSlope of TangentDifferentiation
Tangent Line
A tangent line is a straight line that touches a curve at exactly one point. Imagine drawing a line that lightly grazes the surface of a curve — that’s a tangent line. The point where the tangent meets the curve is called the 'point of tangency'. This line represents the best linear approximation of the curve at that specific point.
At a given point, like where we found on the curve of the function, the tangent line will have the same direction as the curve does at that point. This means the slope of the tangent line is the same as the instantaneous rate of change of the function at this point. Understanding this can help us explore how functions behave at specific values and can aid in prediction and analysis even in practical terms.
At a given point, like where we found on the curve of the function, the tangent line will have the same direction as the curve does at that point. This means the slope of the tangent line is the same as the instantaneous rate of change of the function at this point. Understanding this can help us explore how functions behave at specific values and can aid in prediction and analysis even in practical terms.
When we solve problems involving tangent lines, often we are interested in finding this point and its corresponding slope, which brings us to the concept of differentiation.
Slope of Tangent
The slope of a tangent line is essentially the steepness of the line at a specific point on the curve. Think of it just like you would if you were describing how steep a hill is. In mathematical terms, the slope represents the 'instantaneous rate of change' of the function at a particular point.
The slope is calculated using the derivative of the function. For instance, in our problem, we calculated it using the derivative of the function. When you have a function, like this quadratic one, its derivative, found through differentiation, gives us a formula for the slope at any point.
In the exercise given, the derivative of the function found is used to determine the slope of the tangent line. At the specific x-value provided — which was 2 — we evaluated the derivative to find that the specific slope there was 7.
The slope is calculated using the derivative of the function. For instance, in our problem, we calculated it using the derivative of the function. When you have a function, like this quadratic one, its derivative, found through differentiation, gives us a formula for the slope at any point.
In the exercise given, the derivative of the function found is used to determine the slope of the tangent line. At the specific x-value provided — which was 2 — we evaluated the derivative to find that the specific slope there was 7.
- This slope means that for each unit increase in x, y increases by 7 units at that point of tangency.
- The formula allows us to find the slope at any point easily, which is a powerful tool when we need to understand a function's behavior deeply.
Differentiation
Differentiation is a key concept in calculus that provides us with the derivative of a function. By differentiating, we can find how a function is changing at any given point, which is crucial for tasks like finding the slope of a tangent line.
It's a process that takes a function and evaluates the 'instantaneous rate of change'. When you differentiate an equation, you apply rules like the power rule, product rule, and others to simplify the expression into its derivative form.
In our exercise, we differentiated the quadratic function to get its derivative, which told us how the slope of the tangent line changes as we move along the curve. The power rule was particularly useful here. For a function of the form \( ax^n \), the derivative is \( nax^{n-1} \). This allows us to compute the slope formula for our given function which was \( f(x) = x^2 + 3x + 2 \), resulting in \( f'(x) = 2x + 3 \).
Through repeated practice and recognition of these patterns, differentiation becomes a straightforward tool, empowering you to tackle a range of problems involving rates of change and tangents. Understanding differentiation is essential for grasping how functions behave and how we model real-world phenomena.
It's a process that takes a function and evaluates the 'instantaneous rate of change'. When you differentiate an equation, you apply rules like the power rule, product rule, and others to simplify the expression into its derivative form.
In our exercise, we differentiated the quadratic function to get its derivative, which told us how the slope of the tangent line changes as we move along the curve. The power rule was particularly useful here. For a function of the form \( ax^n \), the derivative is \( nax^{n-1} \). This allows us to compute the slope formula for our given function which was \( f(x) = x^2 + 3x + 2 \), resulting in \( f'(x) = 2x + 3 \).
Through repeated practice and recognition of these patterns, differentiation becomes a straightforward tool, empowering you to tackle a range of problems involving rates of change and tangents. Understanding differentiation is essential for grasping how functions behave and how we model real-world phenomena.
Other exercises in this chapter
Problem 15
Find \(f^{\prime}(x)\). $$f(x)=\sin ^{2} x+\cos ^{2} x$$
View solution Problem 15
$$\text { Find } f^{\prime}(x)$$. $$f(x)=\frac{(2 \sqrt{x}+1)(x-1)}{x+3}$$
View solution Problem 16
Find \(f^{\prime}(x)\) $$f(x)=\tan \sqrt{x}$$
View solution Problem 16
$$\text { Find } f^{\prime}(x)$$ $$f(x)=a x^{3}+b x^{2}+c x+d \quad(a, b, c, d \text { constant })$$
View solution