The linear approximation to (1/sqrt(9-x)) at x=0. What is y?
the linear approximation is just the tangent line at the given point. Within a suitably small interval, the line approximates the curve to any desired accuracy.
So,
y = 1/√(9-x)
y' = 1 / 2(9-x)^(3/2)
at (0,1/3), y' = 1 / 6√3
So, now we have a point and a slope. The line is
y - 1/3 = (1/6√3)(x-0)
To find the linear approximation to the function (1/√(9-x)) at x=0, we can use the formula for the linear approximation:
y = f(a) + f'(a) * (x - a),
where f(a) represents the value of the function at the point a, f'(a) is the derivative of the function at the point a, x is the point at which we want to approximate the function, and y is the approximation.
First, let's find the derivative of the function (1/√(9-x)). To do this, we can use the chain rule.
Let u = 9-x, so the function becomes f(u) = 1/√u.
Using the chain rule, the derivative of f(u) with respect to x is:
f'(x) = f'(u) * u',
where f'(u) represents the derivative of f(u) with respect to u, and u' is the derivative of u with respect to x.
The derivative of f(u) with respect to u can be found using the power rule:
f'(u) = -1/(2√u^3).
The derivative of u with respect to x is simply -1, since the derivative of x is 1.
Now, substituting these values back into the chain rule formula, we have:
f'(x) = (-1/(2√u^3)) * (-1)
= 1/(2√u^3),
where u = 9-x.
Next, we need to find the value of the function and its derivative at the point x=0.
f(0) = 1/√9
= 1/3,
and
f'(0) = 1/(2√9^3)
= 1/(2*27)
= 1/54.
Finally, we can substitute these values into the linear approximation formula:
y = f(0) + f'(0) * (x - 0)
= (1/3) + (1/54) * x.
Therefore, the linear approximation to the function (1/√(9-x)) at x=0 is y = (1/3) + (1/54) * x.