a) To find the derivative of the function f(x) = -3/x at x = 2 using the definition of derivative, we can start by applying the limit definition of the derivative:
f'(x) = lim(h->0) [(f(x + h) - f(x))/h]
Substituting the given function:
f'(x) = lim(h->0) [(-3/(x + h) - (-3/x))/h]
Simplifying the expression:
f'(x) = lim(h->0) [-3(x - (x + h))/(x(x + h)h)]
f'(x) = lim(h->0) [-3(-h)/(x(x + h)h)]
f'(x) = lim(h->0) [3/(x(x + h))]
Now, substitute x = 2 into the expression:
f'(2) = lim(h->0) [3/(2(2 + h))]
Simplifying further:
f'(2) = lim(h->0) [3/(2(2 + h))]
= 3/(2(2)) (since h -> 0, we can substitute h with 0 in the denominator)
= 3/4
Therefore, the derivative of the function f(x) = -3/x at x = 2 is f'(2) = 3/4.
b) To find the equation of the tangent line at x = 2, we can use the point-slope form of a linear equation. We already have the slope, which is the derivative f'(2) = 3/4, and we need a point on the line. We can use the point (2, f(2)).
Substituting x = 2 into the original function:
f(2) = -3/2 = -1.5
So, the point on the line is (2, -1.5).
Using the point-slope form of a linear equation:
y - y1 = m(x - x1)
Substituting the values:
y - (-1.5) = (3/4)(x - 2)
Simplifying:
y + 1.5 = (3/4)(x - 2)
y = (3/4)x - (3/2) - (3/2)
y = (3/4)x - 3/2
Therefore, the equation of the tangent line at x = 2 is y = (3/4)x - 3/2.