Question

In Exercises 35 and \(36,\) find the slope 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

The slope of the tangent line to the function \(y = e^{3x}\) at the point (0,1) is 3, and the slope of the tangent line to the function \(y = e^{-3x}\) at the point (0,1) is -3.

Step-by-step solution

Differentiate the given function

Let's start by differentiating the provided function \(y = e^{3x}\). The derivative of \(e^{3x}\) is \(3e^{3x}\). Therefore, the derivative of the function \(y = e^{3x}\) is \(y' = 3e^{3x}\).

Evaluate the derivative at the given point

Next, evaluate the derivative at the given point (0,1). When \(x=0\), the derivative \(y' = 3e^{3*0}\) simplifies to \(y' = 3e^0\), which further simplifies to \(y' = 3\).

Repeat for the second function

We continue by differentiating the function \(y = e^{-3x}\). The derivative of \(e^{-3x}\) is \(-3e^{-3x}\). Therefore, the derivative of the function \(y = e^{-3x}\) is \(y' = -3e^{-3x}\).

Evaluate the second derivative at the given point

Evaluate the derivative at the given point (0,1). When \(x = 0\), the derivative \(y' = -3e^{-3*0}\) simplifies to \(y' = -3e^0\), which in turn simplifies to \(y' = -3\).