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.