Find a linear approximation of the function f(x)=(1+x)^(1/4) at a=1, and use it to approximate the numbers (.95)^(1/4) and (1.1)^(1/4).
Round your answers to the nearest thousandth
Cheers in advance!
Any curve can be approximated by a straight line, in a small enough interval. So, we want the tangent line at x=0, which will be very close to the curve, if we stay close enough to x=0.
y=(1+x)^(1/4)
y(0) = 1
y' = 1/4 * (1+x)^(-3/4)
= 1/[4(1+x)^3]
y'(0) = 1/4
So, the line y = x/4 + 1 is tangent to f(x) at x=0
let x = -0.05
y = -.0125 + 1 = 0.9875
If x = .1
y = .025 + 1 = 1.025
Just to check,
.95^(1/4) = 0.9873
1.1^(1/4) = 1.0241
Since |.95 - 1| < |1.1 - 1| the approximation is better
To find a linear approximation for the function f(x) = (1+x)^(1/4) at a=1, we start by finding the derivative of f(x) with respect to x.
First, rewrite the function as f(x) = (1+x)^(1/4) = (1+x)^(1/4 * 1).
Using the chain rule, the derivative of f(x) is found by multiplying the derivative of the outer function (1+x)^(1/4) with the derivative of the inner function (1+x), which is 1.
So, the derivative of f(x) is f'(x) = (1/4) * (1+x)^(-3/4) * 1.
Next, evaluate f'(1) to find the slope of the tangent line at x=1.
f'(1) = (1/4) * (1+1)^(-3/4) = (1/4) * 2^(-3/4) = (1/4) * (2^(1/4) / 2) = (1/8) * (2^(1/4)).
Therefore, the slope of the tangent line at x=1 is (1/8) * (2^(1/4)).
The linear approximation of the function f(x) at x=1 can be expressed as:
L(x) = f(1) + f'(1) * (x-1).
Substituting in f(1) = (1+1)^(1/4) = 2^(1/4), we have:
L(x) = 2^(1/4) + (1/8) * (2^(1/4)) * (x-1).
Using this linear approximation, we can approximate the numbers (0.95)^(1/4) and (1.1)^(1/4).
To approximate (0.95)^(1/4), we substitute x=0.95 into our linear approximation:
L(0.95) = 2^(1/4) + (1/8) * (2^(1/4)) * (0.95-1).
= 2^(1/4) + (1/8) * (2^(1/4)) * (-0.05).
= 2^(1/4) - (1/40) * (2^(1/4)).
Approximating this value to the nearest thousandth gives:
L(0.95) ≈ 2^(1/4) - (1/40) * (2^(1/4)) ≈ 1.155.
To approximate (1.1)^(1/4), we substitute x=1.1 into our linear approximation:
L(1.1) = 2^(1/4) + (1/8) * (2^(1/4)) * (1.1-1).
= 2^(1/4) + (1/8) * (2^(1/4)) * (0.1).
= 2^(1/4) + (1/80) * (2^(1/4)).
Approximating this value to the nearest thousandth gives:
L(1.1) ≈ 2^(1/4) + (1/80) * (2^(1/4)) ≈ 1.171.
Therefore, the linear approximations of (0.95)^(1/4) and (1.1)^(1/4) are approximately 1.155 and 1.171, respectively, rounded to the nearest thousandth.
To find a linear approximation of a function at a particular point, we first need to find the equation of the tangent line to the function at that point. The equation of a tangent line can be written in slope-intercept form, y = mx + b, where m is the slope of the tangent line and b is the y-intercept.
To find the slope of the tangent line, we can use the derivative of the function. Let's start by finding the derivative of f(x):
f(x) = (1 + x)^(1/4)
Taking the derivative using the power rule, we get:
f'(x) = (1/4)(1 + x)^(-3/4)
Now, to find the slope of the tangent line at a = 1, we substitute x = 1 into the derivative:
m = f'(1) = (1/4)(1 + 1)^(-3/4) = (1/2)(1)^(−3/4) = (1/2)(1) = 1/2
So, the slope of the tangent line at x = 1 is 1/2.
Next, to find the y-intercept, we substitute the point (1, f(1)) into the equation of the tangent line. In this case, we can substitute (1, f(1)) = (1, (1 + 1)^(1/4)):
y = mx + b
f(1) = (1/2)(1) + b
b = f(1) - (1/2)
b = (1 + 1)^(1/4) - 1/2
So, the equation of the tangent line is:
y = (1/2)x + [(1 + 1)^(1/4) - 1/2]
Now, we can use this linear approximation to estimate the values of (.95)^(1/4) and (1.1)^(1/4). Let's substitute these values into the linear equation:
For (.95)^(1/4):
x = .95
y = (1/2)(.95) + [(1 + 1)^(1/4) - 1/2]
y = 0.475 + [2^(1/4) - 1/2]
y = 0.475 + 0.189 - 0.5
y ≈ 0.164
For (1.1)^(1/4):
x = 1.1
y = (1/2)(1.1) + [(1 + 1)^(1/4) - 1/2]
y = 0.55 + [2^(1/4) - 1/2]
y = 0.55 + 0.189 - 0.5
y ≈ 0.239
Therefore, the linear approximations of (.95)^(1/4) and (1.1)^(1/4) are approximately 0.164 and 0.239, respectively, rounded to the nearest thousandth.