Question

Find an equation of the tangent line to the graph of \(f(x)=4 x e^{\left(x^{2}-1\right)}\) at the point where \(x=-1\). Graph both the function and the tangent line.

Short answer

The tangent line at \(x=-1\) is \(y=12x+8\).

Step-by-step solution

Find the value of the function at the given point

First, we need to evaluate the function at the point where \(x = -1\) to find the y-coordinate of the tangent point. Substitute \(x = -1\) into \(f(x) = 4xe^{(x^2-1)}\): \[ f(-1) = 4(-1)e^{((-1)^2-1)} = -4e^{(1-1)} = -4e^0 = -4 \] Thus, the point on the graph is \((-1, -4)\).

Find the derivative of the function

We need to find \(f'(x)\), the derivative of \(f(x)\), to get the slope of the tangent line. The function is \(f(x) = 4xe^{(x^2-1)}\). This is a product of two functions: \(u(x) = 4x\) and \(v(x) = e^{(x^2-1)}\). We use the product rule: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Where:- \(u'(x) = 4\)- \(v(x) = e^{(x^2-1)}\) and \(v'(x) = e^{(x^2-1)} \cdot (2x)\). Thus,\[ f'(x) = 4e^{(x^2-1)} + 4x \cdot 2x \cdot e^{(x^2-1)} \] Simplifying,\[ f'(x) = 4e^{(x^2-1)} + 8x^2e^{(x^2-1)} = (4 + 8x^2)e^{(x^2-1)} \]

Evaluate the derivative at the given point

To find the slope of the tangent line at \(x = -1\), substitute \(x = -1\) into \(f'(x) = (4 + 8x^2)e^{(x^2-1)}\):\[ f'(-1) = (4 + 8(-1)^2)e^{((-1)^2-1)} = (4 + 8)e^{(1-1)} = 12e^0 = 12 \] The slope of the tangent line at \(x = -1\) is 12.

Find the equation of the tangent line

Using the point-slope formula \(y - y_1 = m(x - x_1)\), where \(m = 12\), \(x_1 = -1\), and \(y_1 = -4\), we find the equation of the tangent line:\[ y - (-4) = 12(x - (-1)) \]\[ y + 4 = 12(x + 1) \]Simplifying,\[ y = 12x + 12 - 4 \]\[ y = 12x + 8 \] The equation of the tangent line is \(y = 12x + 8\).

Graphing the function and the tangent line

To plot the graph, we can use graphing software to visualize both the function and the tangent line. The function \(f(x) = 4xe^{(x^2-1)}\) has a complicated shape, but we plot it over a reasonable range to observe its behavior around \(x = -1\). The tangent line \(y = 12x + 8\) is a straight line intersecting the graph of the function exactly at the point \((-1, -4)\).

Key concepts

Tangent Line

A tangent line is a line that touches a curve at a single point without crossing it. In calculus, tangent lines are vital because they represent the instantaneous rate of change of the function at a given point.
To find a tangent line to a function like \( f(x) = 4x e^{(x^2-1)} \) at point \( x = -1 \), follow these steps:

• First, evaluate the function at \( x = -1 \) to find the point of contact. In this exercise, \( f(-1) = -4 \), so the point is \((-1, -4)\).
• Next, determine the slope of the tangent line using the derivative of the function at \(x = -1\).
• Use the point-slope form of a line equation \(y - y_1 = m(x - x_1)\) to find the equation of the tangent line, where \(m\) is the slope and \((x_1, y_1)\) is the point of tangency.

The tangent line in this example is \(y = 12x + 8\), meaning it has a slope of 12.

Derivative

The derivative of a function represents its rate of change. For a function \( f(x) \), its derivative, denoted \( f'(x) \), gives the slope of the tangent line to the curve at any point \( x \).
Finding the derivative is crucial for understanding how a function behaves and predicting changes.
In our specific exercise, the task is to find \( f'(x) \) of the function \( f(x) = 4x e^{(x^2-1)} \):

• Identifying \( u(x) = 4x \) and \( v(x) = e^{(x^2-1)} \), you can apply the product rule to find \( f'(x) \).
• The derivative \( u'(x) = 4 \), and \( v'(x) = e^{(x^2-1)} \cdot 2x \).
• Thus, the derivative is \( f'(x) = 4e^{(x^2-1)} + 8x^2e^{(x^2-1)} \).

The derivative lets us ascertain the behavior of the function around \( x = -1 \), leading to a slope of 12 for the tangent line at this point.

Product Rule

The product rule is a technique for finding the derivative of a function that is the product of two functions. Given a function \( f(x) = u(x)v(x) \), the product rule states:\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
When you have a function like \( f(x) = 4x e^{(x^2-1)} \), you identify:

• \( u(x) = 4x \) and \( v(x) = e^{(x^2-1)} \)
• \( u'(x) = 4 \)
• \( v'(x) = e^{(x^2-1)} \cdot 2x \)

Substitute these into the product rule formula, and you'll get:

• \( f'(x) = 4e^{(x^2-1)} + 4x \cdot 2x \cdot e^{(x^2-1)} \)
• Simplify to \( f'(x) = (4 + 8x^2)e^{(x^2-1)} \)

The product rule allows us to accurately calculate the derivative, which leads to identifying the slope at specific points on the function.

Function Evaluation

Function evaluation involves calculating the value of a function at a specific point, which is essential for finding the exact point where the tangent line touches the curve.
At \( x = -1 \), we need to find \( f(-1) \) in the function: \( f(x) = 4x e^{(x^2-1)} \).

• Substitute \( x = -1 \) into the function.
• This becomes \( f(-1) = 4(-1)e^{((-1)^2-1)} = -4e^{(0)} = -4 \).

Through function evaluation, we find the specific point \((-1, -4)\) on the graph. This provides the necessary coordinates for the equation of the tangent line.
Knowing how to evaluate functions correctly ensures precision when working with graphs and solving calculus problems like finding tangent lines.