The linearization at a=0 to sqrt(1+2x) is A+Bx where A is____ and where B is _____?
A=?
B=?
Ty
the linearization is just finding a straight line that is close to the curve at the given point. That is just the tangent line. So, since when
y = √(1+2x),
y' = 1/√(1+2x)
y(0) = 1
y'(0) = 1
and the point-slope form for the line is
y-1 = 1(x-0)
y = x+1
A=1 and B=1
Check the graphs at
http://www.wolframalpha.com/input/?i=plot+y+%3D+%E2%88%9A%281%2B2x%29+and+y%3Dx%2B1+for+-.1+%3C+x+%3C+.1
You can see that at (0,1) the line and the curve are very close.
To find the linearization at a=0 for the function f(x) = sqrt(1 + 2x), we need to determine the values of A and B in the linearization A + Bx.
We start by finding the first derivative of the function:
f'(x) = d/dx [sqrt(1 + 2x)]
To accomplish this, we can use the chain rule.
Let u = 1 + 2x
Then, y = sqrt(u)
We can rewrite y = sqrt(u) as y = u^(1/2)
Using the chain rule, we have:
dy/dx = (dy/du) * (du/dx)
= (1/2) * (1/sqrt(u)) * (d(1 + 2x)/dx)
= (1/2) * (1/sqrt(1 + 2x)) * 2
= 1/sqrt(1 + 2x)
Now, let's evaluate the derivative at a=0:
f'(0) = 1/sqrt(1 + 2(0))
= 1/sqrt(1)
= 1
Now we can find A and B in the linearization:
A = f(0)
A = sqrt(1 + 2(0))
A = sqrt(1)
A = 1
B = f'(0)
B = 1
Therefore, in the linearization A + Bx, A = 1 and B = 1.
To find the linearization at a point, you can use the linearization formula:
L(x) = f(a) + f'(a)(x - a)
Where:
- L(x) is the linearization of the function f(x) at point a.
- f(a) is the value of the function f(x) at point a.
- f'(a) is the derivative of the function f(x) evaluated at point a.
- (x - a) is the difference between the input x and the value a.
In this case, the function is f(x) = sqrt(1 + 2x), and the point of interest is a = 0.
Step 1: Determine the value of f(a)
Evaluate the function at a = 0:
f(0) = sqrt(1 + 2(0))
f(0) = sqrt(1)
f(0) = 1
So, f(a) = 1.
Step 2: Determine the value of f'(a)
Take the derivative of f(x) with respect to x:
f'(x) = d/dx[sqrt(1 + 2x)]
f'(x) = (2/(2*sqrt(1 + 2x)) = 1/sqrt(1 + 2x)
Evaluate the derivative at a = 0:
f'(0) = 1/sqrt(1 + 2(0))
f'(0) = 1/sqrt(1)
f'(0) = 1/1
f'(0) = 1
So, f'(a) = 1.
Step 3: Substitute the values into the linearization formula
L(x) = f(a) + f'(a)(x - a)
L(x) = 1 + 1(x - 0)
L(x) = 1 + x
Therefore, the linearization of the function sqrt(1 + 2x) at a = 0 is A + Bx, where A = 1 and B = 1.