what is the linear approximation of f(x)=x^3-x at x=2
dy/dx = 3 x^2 - 1
at x = 2, dy/dx = 3*4-1 = 11
so m, the slope, = 11
at x = 2, y = 8 - 2 = 6
so
we need a line of form y = m x + b
with slope m = 11
that goes through the point (2,6)
To find the linear approximation of a function at a specific point, we can use the derivative of the function. The linear approximation is given by:
L(x) = f(a) + f'(a)(x - a)
where f(a) represents the value of f(x) at the point of interest, a, and f'(a) is the derivative of f(x) evaluated at a.
First, let's find the value of f(a) at x = 2:
f(2) = (2)^3 - 2 = 8 - 2 = 6
Next, we need to find the derivative of f(x). For f(x) = x^3 - x, the derivative is:
f'(x) = 3x^2 - 1
Now, evaluate f'(a) at x = 2:
f'(2) = 3(2)^2 - 1 = 12 - 1 = 11
Now, we have all the information we need to find the linear approximation L(x) at x = 2:
L(x) = f(2) + f'(2)(x - 2)
Plugging in the values we found earlier:
L(x) = 6 + 11(x - 2)
Simplifying this expression, we get the linear approximation of f(x) = x^3 - x at x = 2:
L(x) = 11x - 16