Question

find the equation of the line tangent to the graph of f at the indicated value of x. f(x)=1 - ln x; x = 1

Explanation:

Step1: Find the function value at x = 1

Substitute x = 1 into f(x): $f(1)=1-\ln(1)=1 - 0=1$.

Step2: Find the derivative of the function

The derivative of $f(x)=1-\ln x$ is $f^\prime(x)=-\frac{1}{x}$.

Step3: Find the slope of the tangent line at x = 1

Substitute x = 1 into $f^\prime(x)$: $f^\prime(1)=-\frac{1}{1}=-1$.

Step4: Use the point - slope form to find the tangent line equation

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,1)$ and $m=-1$.
$y - 1=-1(x - 1)$.
Simplify to get $y=-x + 2$.

Answer:

$y=-x + 2$