(a) Find the tangent line approximation for sqrt(9+x) near x=0.
(b)Find a formula for the error, E(x),in the tangent line approximation found in part (a).
For part A, is the answer y=3. For part b, I don't understand , any help would be greatly appreciated.
(a) What they are calling the "tangent line approximation" is usually called the first two terms of a Taylor series.
If f(x) = sqrt(9 + x), for small values of x, f(x) is very close to
f(0) + df/dx(0) *x
= 3 + (x/6)
(b) The difference between the actual value of sqrt(x+9) and the tangent line approximation is
E(x) = sqrt(x+9) -3 - x/6
find tangent line approximation of 2 over x+9
(a) To find the tangent line approximation for sqrt(9+x) near x=0, we will use the concept of linear approximation. The formula for the tangent line approximation (or linear approximation) is given by:
y ≈ f(a) + f'(a)(x - a)
where f(a) is the value of the function at the point a, f'(a) is the derivative of the function at the point a, x is the variable, and a is the point around which we are approximating.
In this case, f(x) = sqrt(9+x), a = 0, and we need to find f'(a).
To find f'(x), we can use the power rule for differentiation.
f'(x) = d/dx(sqrt(9+x))
= 1/2 * (9+x)^(-1/2)
= 1/(2sqrt(9+x))
Substituting a = 0 and f'(a) into the tangent line approximation formula, we get:
y ≈ f(0) + f'(0)(x - 0)
≈ sqrt(9+0) + 1/(2sqrt(9+0)) * (x - 0)
≈ 3 + 1/(6) * x
≈ 3 + (1/6)x
Therefore, the tangent line approximation for sqrt(9+x) near x=0 is y = 3 + (1/6)x.
(b) To find the formula for the error, E(x), we need to take into account the difference between the actual function (sqrt(9+x)) and its tangent line approximation.
E(x) = f(x) - y
Substituting the values of f(x) and y, we get:
E(x) = sqrt(9+x) - (3 + (1/6)x)
Simplifying further, we have:
E(x) = sqrt(9+x) - 3 - (1/6)x
This formula calculates the error between the actual function and its tangent line approximation.
To find the tangent line approximation for sqrt(9+x) near x=0, we need to use the concept of the derivative.
(a) Tangent line approximation:
1. Take the derivative of the given function sqrt(9+x) with respect to x.
f'(x) = (1/2)*(9+x)^(-1/2)
2. Evaluate the derivative at x=0 to find the slope of the tangent line.
f'(0) = (1/2)*(9+0)^(-1/2) = (1/2)*(9)^(-1/2) = 1/6
3. Use the point-slope form of a line equation to find the equation of the tangent line.
y - y1 = m(x - x1), where (x1, y1) is the given point (0, sqrt(9+0)) = (0, 3) and m is the slope computed in step 2.
y - 3 = (1/6)(x - 0)
y = (1/6)x + 3
So, the tangent line approximation for sqrt(9+x) near x=0 is y = (1/6)x + 3.
(b) Formula for the error, E(x), in the tangent line approximation:
The error in the tangent line approximation is the difference between the actual function value and the value given by the tangent line. We can express this as:
E(x) = f(x) - T(x)
where E(x) is the error, f(x) is the value of the original function, and T(x) is the value given by the tangent line.
In this case, f(x) = sqrt(9+x) and T(x) = (1/6)x + 3 (as found in part (a)).
E(x) = sqrt(9+x) - [(1/6)x + 3]
Simplifying this expression further will yield the formula for the error, E(x), in the tangent line approximation for sqrt(9+x) near x=0.