Find the domain and range of the relation. Use interval notation where appropriate.
1 answer:
You might be interested in
Answer:
a) f(x) = 4 - x²
The linear approximation of the function at a=1 is
L(x) = 5 - 2x at a = 1
b) The graph of the function and the linear approximation at that point is attached to this solution.
The curve represent the real function,
f(x) = 4 - x²
The straight line represents the linear approximation of the function at a=1.
L(x) = 5 - 2x
The curve and the function evidently cross paths at x=1 and understandably so.
c) Using the linear approximation obtained at a = 1.
f(1.1) = 2.8
Using the actual function, the actual value of f(1.1) = 2.79
d) Percent error = 0.358%
Step-by-step explanation:
f(x) = 4 - x²
a) The linear approximation of the function at the given point is given as
L(x) = f(a) + f'(a) [x - a]
f(x) = 4 - x²
a = 1
f(a) = 4 - 1² = 3
f'(x) = -2x
f'(a) = -2(1) = -2
L(x) = f(a) + f'(a) [x - a]
L(x) = 3 + (-2)(x - 1)
L(x) = 3 -2x + 2
L(x) = 5 - 2x
L(x) = -2x + 5
f(x) = 4 - x²
L(x) = 5 - 2x at a = 1
b) The graph of the function and the linear approximation at that point is attached to this solution.
The curve represent the real function,
f(x) = 4 - x²
The straight line represents the linear approximation of the function at a=1.
L(x) = 5 - 2x
The curve and the function evidently cross paths at x=1 and understandably so.
c) Use the linear approx. to estimate the given fxn value.
f(1.1)
L(x) = 5 - 2x
L(1.1) = 5 - 2(1.1) = 2.8
Using the function, the actual value of f(1.1) = 4 - 1.1² = 2.79
d) Compute the percent error in your approximation, 100*Iapprox-exactI/IexactI, where the exact value is given by a calculator
Percent error
= 100% × (|approx - exact|)/exact
Approximated value = 2.8
Exact value = 2.79
Percent error = 100% × (2.8-2.79)/2.79
Percent error = 0.358%
Hope this Helps!!!
