Answer:
tangent line to the graph of f at x=2 is:
To find the equation of the tangent line, we first need to find the slope of the tangent line. The slope of the tangent line is equal to the derivative of f(x) evaluated at x=2.
f'(x) = 9 - (8/x^3)
f'(2) = 9 - (8/2^3) = 9 - (8/8) = 9 - 1 = 8
So the slope of the tangent line is 8.
Next, we need to find the y-coordinate of the point on the graph of f at x=2. We can plug x=2 into the equation for f(x).
f(2) = 9(2) + (4/2^2) = 18 + 1 = 19
So the point on the graph of f at x=2 is (2, 19).
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 write the equation of the tangent line as:
y - 19 = 8(x - 2)
Simplifying,
y - 19 = 8x - 16
y = 8x + 3
Therefore, the equation of the tangent line to the graph of f at x=2 is y = 8x + 3.
Explanation: