Questions: Given the function f(x)=-4x-7+6e^x Determine the equation of the tangent line at the point (0,-1).

Transcript text: Given the function \[ f(x)=-4 x-7+6 e^{x} \] Determine the equation of the tangent line at the point $(0,-1)$. $\square$

Solution

Solution Steps

To find the equation of the tangent line at a given point, we need to:

Compute the derivative of the function $ f(x) $ to find the slope of the tangent line.
Evaluate the derivative at the given point $ x = 0 $ to get the slope at that point.
Use the point-slope form of the equation of a line, $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $ (x_1, y_1) $ is the given point.

Step 1: Define the Function

We start with the function given by \[ f(x) = -4x - 7 + 6e^{x}. \]

Step 2: Compute the Derivative

Next, we compute the derivative of the function: \[ f'(x) = 6e^{x} - 4. \]

Step 3: Evaluate the Derivative at $ x = 0 $

We evaluate the derivative at the point $ x = 0 $: \[ f'(0) = 6e^{0} - 4 = 6 - 4 = 2. \] Thus, the slope of the tangent line at the point $ (0, -1) $ is $ 2 $.

Step 4: Use the Point-Slope Form

Using the point-slope form of the equation of a line, we have: \[ y - y_1 = m(x - x_1), \] where $ m = 2 $, $ (x_1, y_1) = (0, -1) $. Substituting these values gives: \[ y - (-1) = 2(x - 0). \] This simplifies to: \[ y + 1 = 2x. \] Rearranging this, we find: \[ y = 2x - 1. \]