let f(x)=(x-1)^2, g(x)=e^(-2x), and h(x)=1+ ln(1-2x)
find linearization of f, g, and h at a=0
Lf(x)=
Lg(x)=
Lh(x)=
graph f,g, and h, and their linear app. For which function is the linear app. best? for which is the worst? explain.
the linear app. appears to be best for the function ______ since it is closer to _____ for a larger domain than it is to _____ . the app. looks worst for _____ since _____ moves away from L faster than _____ do.
the linear app. appears to be best for the function ___f___ since it is closer to ___f__ for a larger domain than it is to _g and h____ . the app. looks worst for ___h__ since _h____ moves away from L faster than __f and g___ do.
Lf(x)= 1-2x
Lg(x)= 1-2x
Lh(x)= 1-2x
the linear app. appears to be best for the function ___f___ since it is closer to _____ for a larger domain than it is to _____ . the app. looks worst for ___h__ since _____ moves away from L faster than _____ do.
i couldn't figure out the rest!!!
Lf(x) = f(a) + f'(a)(x - a)
= f(0) + f'(0)(x - 0)
= (0 - 1)^2 + 2(0 - 1)(x - 0)
= x^2 - 2x
Lg(x) = g(a) + g'(a)(x - a)
= g(0) + g'(0)(x - 0)
= e^(-2(0)) + (-2e^(-2(0)))(x - 0)
= 1 - 2x
Lh(x) = h(a) + h'(a)(x - a)
= h(0) + h'(0)(x - 0)
= 1 + ln(1 - 2(0)) + (-2/(1 - 2(0)))(x - 0)
= 1 + ln(1) - 2x
= 1 - 2x
To graph these functions and their linear approximations, we can use a graphing calculator or software.
The linear approximation appears to be best for the function f(x) since it is closer to f(x) for a larger domain than it is to g(x) or h(x). The linear approximation looks worst for the function h(x) since h(x) moves away from Lh(x) faster than f(x) or g(x) do.
To find the linearization of a function at a specific point, we can use the Taylor series expansion. The linearization Lf(x) of f(x) at a=0 is given by the first two terms of the Taylor series:
Lf(x) = f(0) + f'(0)(x-0)
To find the linearization of g(x) and h(x) at a=0, we will follow the same process using their corresponding Taylor series expansions.
First, let's find the linearization of f(x):
f(x) = (x-1)^2
Taking the derivative:
f'(x) = 2(x-1)
Substituting x=0:
f'(0) = 2(0-1) = -2
Now we can find the linearization:
Lf(x) = f(0) + f'(0)(x-0)
= (0-1)^2 + (-2)(x-0)
= 1 - 2x
Now, let's find the linearization of g(x):
g(x) = e^(-2x)
Taking the derivative:
g'(x) = -2e^(-2x)
Substituting x=0:
g'(0) = -2e^0 = -2(1) = -2
Now we can find the linearization:
Lg(x) = g(0) + g'(0)(x-0)
= e^(-2(0)) + (-2)(x-0)
= 1 - 2x
Finally, let's find the linearization of h(x):
h(x) = 1 + ln(1-2x)
Taking the derivative:
h'(x) = -2/(1-2x)
Substituting x=0:
h'(0) = -2/(1-2(0)) = -2/1 = -2
Now we can find the linearization:
Lh(x) = h(0) + h'(0)(x-0)
= 1 + ln(1-2(0)) + (-2)(x-0)
= 1 - 2x
To compare the accuracy of the linear approximations, it will be helpful to graph f(x), g(x), h(x), Lf(x), Lg(x), and Lh(x).
The linear approximations Lf(x), Lg(x), and Lh(x) will be lines while the original functions f(x), g(x), and h(x) will be curves.
The graph of f(x) will be a parabola, g(x) will be a decreasing exponential curve, and h(x) will be a logarithmic curve.
We can visually compare the linear approximations to the original functions to determine which linear approximation is the best and which one is the worst.
For the function f(x), the linear approximation Lf(x) given by 1 - 2x will be the best because, for a larger domain, it will be closer to f(x) compared to g(x) and h(x).
For the function g(x), the linear approximation Lg(x) given by 1 - 2x will be the worst because g(x) will move away from Lg(x) faster than f(x) and h(x) do.
For the function h(x), the linear approximation Lh(x) given by 1 - 2x will be closer to h(x) for a larger domain compared to g(x), but not as close as Lf(x) is to f(x).
To find the linearization of a function at a specific point, we need to find the equation of the tangent line to the graph of the function at that point.
1. Linearization of f(x):
To find the linearization Lf(x) of f(x) at a=0, we first need to find the derivative of f(x) with respect to x:
f'(x) = 2(x-1)
The slope of the tangent line at x=0 is given by f'(0) = 2(0-1) = -2.
Since we are finding the linearization at a=0, the x-coordinate of the point is 0. We can use the point-slope form of a line to find the equation of the tangent line:
y - f(0) = f'(0)(x - 0)
Now, substitute the values in:
y - f(0) = -2x
y = -2x + 1
Therefore, the linearization of f(x) at a=0 is Lf(x) = -2x + 1.
2. Linearization of g(x):
To find the linearization Lg(x) of g(x) at a=0, we first need to find the derivative of g(x) with respect to x:
g'(x) = -2e^(-2x)
The slope of the tangent line at x=0 is given by g'(0) = -2e^0 = -2.
Using the point-slope form of a line with the point (0, g(0)), we get:
y - g(0) = g'(0)(x - 0)
Now, substitute the values in:
y - 1 = -2x
y = -2x + 1
Therefore, the linearization of g(x) at a=0 is Lg(x) = -2x + 1.
3. Linearization of h(x):
To find the linearization Lh(x) of h(x) at a=0, we need to find the derivative of h(x) with respect to x:
h'(x) = -2/(1 - 2x)
The slope of the tangent line at x=0 is given by h'(0) = -2/(1 - 2*(0)) = -2/1 = -2.
Using the point-slope form of a line with the point (0, h(0)), we get:
y - h(0) = h'(0)(x - 0)
Now, substitute the values in:
y - 1 = -2x
y = -2x + 1
Therefore, the linearization of h(x) at a=0 is Lh(x) = -2x + 1.
Graphing the functions f(x), g(x), h(x), and their linear approximations Lf(x), Lg(x), Lh(x), we can analyze how well the linear approximations fit the functions.
For which function is the linear approximation best?
The linear approximation appears to be best for function g(x) since the graph of g(x) is relatively close to its linear approximation Lg(x) over a larger domain than the other functions.
For which function is the linear approximation worst?
The linear approximation appears to be worst for function h(x) since the graph of h(x) moves away from its linear approximation Lh(x) faster than the graphs of f(x) and g(x) do.