# How to find the equation of the line tangent to the graph of f at the indicated value of x? f(x)= 4 - lnx ; x=1

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To find the equation of the line tangent to the graph of f at x=1, we need to find the derivative of f(x) and evaluate it at x=1.

The derivative of f(x) can be found using the power rule and the chain rule. The derivative of 4 is 0, and the derivative of ln(x) is 1/x. Therefore, the derivative of f(x) is 0 - 1/x.

Evaluating the derivative at x=1, we have 0 - 1/1 = -1.

So, the slope of the tangent line at x=1 is -1.

To find the equation of the line, we need a point on the line. Since the line is tangent to the graph of f at x=1, we can use the point (1, f(1)).

Plugging x=1 into f(x), we have f(1) = 4 - ln(1) = 4 - 0 = 4.

Therefore, the point on the line is (1, 4).

Using the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we can substitute the values:

y - 4 = -1(x - 1)

Simplifying, we have y - 4 = -x + 1.

Rearranging the equation, we get y = -x + 5.

So, the equation of the line tangent to the graph of f at x=1 is y = -x + 5.

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