find the linear approximation to f(x)=1/square root of(x^3+1) at a=2
f(x) = (x^3+1)^(-1/2)
f'(x) = -3/2 x^2 (x^3+1)^(-3/2)
f(2) = 1/3
f'(2) = -2/9
the linear approximation is just the tangent line at x=2, so
y - 1/3 = -2/9 (x-2)
see the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D1%2F%E2%88%9A(x%5E3%2B1),+y+%3D+-2%2F9+(x-2)+%2B+1%2F3
To find the linear approximation to the function f(x) at a given point a, we use the formula:
L(x) = f(a) + f'(a)(x - a)
where L(x) is the linear approximation, f(a) is the value of the function at the point a, f'(a) is the derivative of the function evaluated at a, and (x - a) represents the difference between the given x value and the point a.
First, let's find f(a) by substituting a = 2 into the function:
f(a) = 1/sqrt(a^3 + 1) = 1/sqrt(2^3 + 1) = 1/sqrt(8 + 1) = 1/sqrt(9) = 1/3
Next, let's find f'(a) by taking the derivative of the function:
f(x) = 1/sqrt(x^3 + 1)
Using the power rule and chain rule, we can find the derivative:
f'(x) = -3x^2/(2(x^3 + 1)^(3/2))
Substituting a = 2 into this derivative equation:
f'(a) = -3(2)^2/(2(2^3 + 1)^(3/2)) = -12/(2(8 + 1)^(3/2)) = -12/(2(9)^(3/2)) = -12/(2(27)) = -12/54 = -2/9
Now we can substitute f(a) = 1/3 and f'(a) = -2/9 into the linear approximation formula:
L(x) = f(a) + f'(a)(x - a) = (1/3) + (-2/9)(x - 2)
Therefore, the linear approximation to f(x) at a = 2 is L(x) = (1/3) - (2/9)(x - 2).