Chapter 5: Problem 33
In Exercises 33 and 34 , find an equation of the tangent line to the graph of the function at the point \((0,1)\).(a) \(y=e^{3 x}\) (b) \(y=e^{-3 x}\)
Short Answer
Expert verified
For the function \(y = e^{3x}\) the tangent line is \(y = 3x+1\). For the function \(y = e^{-3x}\) the tangent line is \(y = -3x+1\).
Step by step solution
01
Find the Derivative
To find the slope of a tangent line to the function \(y = e^{3x}\) at a given point, we have to first find the derivative of the function. This is done using the chain rule of differentiation, yielding: \(y'(x) = 3e^{3x}\)
02
Evaluate the Derivative at the Given Point
We substitute \(x = 0\) into \(y'(x)\) to find the slope of the tangent line at the point (0,1). Thus, \(y'(0) = 3e^{3*0} = 3\).
03
Use the Point-Slope form to find the Equation of the Tangent Line
The point-slope form of a linear equation is given by \(y - y1 = m(x - x1)\). Here, \(m\) is the slope of the tangent line, and (x1, y1) is the given point. Substituting these values: \(x1 = 0, y1 = 1\), and \(m = 3\), the equation of the line for (a) becomes: \(y - 1 = 3(x - 0)\). So, the equation of the tangent line is \(y = 3x + 1\).
04
Repeat Steps for Part (b)
The derivative of \(y = e^{-3x}\) is \(y'(x) = -3e^{-3x}\) using the chain rule. Substituting \(x = 0\) we find that \(y'(0) = -3e^{-3*0} = -3\). The equation of the line for (b) becomes: \(y - 1 = -3(x - 0)\). So the equation for this tangent line is \(y = -3x + 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. A composite function is one in which one function is nested inside another. For example, in the function \(y = e^{3x}\), we have the exponential function \(e^u\), where \(u = 3x\). The chain rule helps us differentiate such functions efficiently. To apply the chain rule, we follow these steps:
- Differentiate the outer function with respect to its inner function.
- Multiply by the derivative of the inner function.
Derivative
The derivative of a function represents the rate of change or the slope of the function at any given point. In simple terms, it tells us how steep the function is at a particular point. For the function \(y = e^{3x}\), the derivative was computed as \(y'(x) = 3e^{3x}\). This formula helps us understand how the function grows as \(x\) changes.
To find the exact slope of the tangent line at a specific point, we substitute the point's \(x\)-coordinate into the derivative. In this case, substituting \(x = 0\) into \(y'(x)\) gives us \(3e^{0} = 3\). This means at the point \((0, 1)\), the slope of the tangent line to the graph of \(y = e^{3x}\) is \(3\). A similar process is followed for \(y = e^{-3x}\) to find its slope.
To find the exact slope of the tangent line at a specific point, we substitute the point's \(x\)-coordinate into the derivative. In this case, substituting \(x = 0\) into \(y'(x)\) gives us \(3e^{0} = 3\). This means at the point \((0, 1)\), the slope of the tangent line to the graph of \(y = e^{3x}\) is \(3\). A similar process is followed for \(y = e^{-3x}\) to find its slope.
Exponential Function
Exponential functions are a type of mathematical function where a constant base, such as \(e\), is raised to a variable exponent. These functions are frequently used in real-world situations like compound interest, population growth, and radioactive decay. The function \(y = e^{3x}\) is an example where the constant base \(e\) is raised to the power of \(3x\).
- The steep increase or decrease behaviour of exponential functions is due to the exponent, which leads to rapid growth or decay.
- In our exercise, we see both increasing exponential function \(e^{3x}\) and decreasing exponential function \(e^{-3x}\).
Point-Slope Form
The point-slope form of a linear equation is a method to write the equation of a line if you know the slope and a point on the line. The formula is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, and \((x_1, y_1)\) is a point on the line.
- It is particularly useful for writing equations of tangent lines, as we often know a specific point and the slope at that point from the derivative.
- The formula rearranges to make \(y\) the subject, giving a classic line equation \(y = mx + b\), where \(b\) is the y-intercept.