Asked by Alison
                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!
            
        Round your answers to the nearest thousandth
Cheers in advance!
Answers
                    Answered by
            Steve
            
    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
    
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
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.