STATE THE LINEAR APPROXIMATION FOR: (2.04)raise to power 5
There are lots of possible approximations. If we have
y = x^5, the slope of the tangent line is
y' = 5x^4
At (2,32), the slope is 80
so, if a(x) is the line,
a(x) = 80(x-2)+32
a(2.04) = 80(.04)+32 = 35.2
In fact, 2.04^5 = 35.33
To find the linear approximation for the expression (2.04) raised to the power of 5, we can use the concept of the tangent line.
First, let's define our function: f(x) = x^5. We want to find the approximation of f(2.04).
The linear approximation of a function f(x) at a specific point a can be represented as:
L(x) = f(a) + f'(a)(x - a)
where f'(a) is the derivative of the function evaluated at point a.
Now, let's differentiate f(x) = x^5 with respect to x to find f'(x):
f'(x) = 5x^4
To find the linear approximation of f(2.04), we need to evaluate f(2.04) and f'(2.04).
f(2.04) = (2.04)^5 = 45.83510096
f'(2.04) = 5(2.04)^4 ≈ 82.343424
Next, we substitute the values into the linear approximation formula:
L(x) = f(2.04) + f'(2.04)(x - 2.04)
L(x) = 45.83510096 + 82.343424(x - 2.04)
Therefore, the linear approximation of (2.04)^5 is given by:
L(x) = 45.83510096 + 82.343424(x - 2.04)