To find the equation of the tangent line to the graph of a function at a given point, we must first find the derivative (slope) of the function at that point and then apply this in the point-slope form of the linear equation.
Let's substantiate this:
The function given is f(x) = 8 - ln(x).
1. **Finding the derivative of the function**:
We know that the slope of the tangent line at a given point on the function is the derivative of the function at that point. The derivative of 8 (a constant) is 0 and the derivative of -ln(x) is -1/x. So, the derivative of the function f(x) = 8 - ln(x) is f'(x) = -1/x.
2. **Evaluating the derivative at the given point**:
We substitute x = 1 in f'(x). Hence, f'(1) = -1/1 = -1. This is the slope of the tangent line at x = 1.
3. **Finding the value of the function at the given point**:
Evaluate the function f(x) = 8 - ln(x) at x=1. By substituting x = 1, we find that f(1) = 8 - ln(1) = 8 - 0 = 8.
4. **Deriving the equation of the tangent line using the point-slope form**:
We use the standard point-slope form of the equation of a line y - y1 = m(x - x1), where m is the slope of the line, and (x1, y1) is the given point on the line. Here, the slope m is -1 (which we calculated in step 2), and (x1, y1) is the point (1,8). Substituting these values into the equation, we get y - 8 = -1(x - 1).
So, the equation of the line tangent to the graph of the function f(x) = 8 - ln(x) at x = 1 is y - 8 = -1(x - 1).