Suppose the function f(x)=4/x

a) Use the definition of the derivative to show that f ' (-2) = -1.
b) Write an equation for the line tangent to the graph of f at x = -2.

f(-2) = 4/-2 = -2

f(-2+h) = 4/(-2+h)

f(-2+h)-f(-2) = 4/(-2+h) + 2
= [ 4 -4+2h ] / (-2+h)
= 2 h /(-2+h)

[f(-2+h)-f(-2)]/h = 2/(-2+h)
limit as h -->0 = 2/-2 = -1

b)
slope = m = -1
y = -x + b
goes through (-2, -2)
-2 = 2 + b
b = -4
y = -x -4

what is h?

To solve this problem, let's start by finding the derivative of the function f(x) using the definition of the derivative.

a) Definition of the derivative:
The derivative of a function f(x) at a point x = a is defined as the limit of the difference quotient as h approaches zero:

f ' (a) = lim(h->0) [(f(a + h) - f(a))/h]

In our case, the function is f(x) = 4/x. Therefore, we can substitute this expression into the definition:

f ' (x) = lim(h->0) [(4/(x + h) - 4/x)/h]

Now, let's apply the algebraic manipulations to simplify the expression:

f ' (x) = lim(h->0) [(4x - 4(x + h))/(x(x + h)h)]

f ' (x) = lim(h->0) [(-4h)/(x(x + h)h)]

f ' (x) = lim(h->0) [-4/(x(x + h))]

Now, let's evaluate the limit as h approaches zero:

f ' (x) = -4/(x(x + 0)) = -4/(x²)

Now, we need to find f ' (-2):

f ' (-2) = -4/((-2)²) = -4/4 = -1

Therefore, f ' (-2) = -1.

b) To write the equation of the tangent line at x = -2, we can use the point-slope form of a linear equation:

y - y₁ = m(x - x₁)

where (x₁, y₁) represents the point (-2, f(-2)), and m represents the slope, which is the derivative of the function evaluated at x = -2.

Since f ' (-2) = -1, we have:

m = -1

Now, substitute the values into the equation:

y - f(-2) = -1(x - (-2))

Since f(x) = 4/x, we can substitute the value of f(-2):

y - (-2) = -1(x - (-2))

Simplifying, we get:

y + 2 = -1(x + 2)

y + 2 = -x - 2

Now, rearrange the equation to get it in the slope-intercept form (y = mx + b):

y = -x - 4

Therefore, the equation of the line tangent to the graph of f at x = -2 is y = -x - 4.