find the linear approximation to f(x)=(x+1)^1/3 at x=0 and use the linear approximation to approximate f(0.1)?

Answer 1

I guess your function is:
f(x)=(x+1)^(1/3)
instead of f(x)=(x+1)^1 /3 = x/3 +1/3, right? Please pay attention to the order of operation.
According to Newton's generalized binomial theorem, we have:
f(x)=(x+1)^(1/3) = (1+x)^(1/3)
= 1 + (1/3)x + (1/3)(-2/3)x^2/2! + (1/3)(-2/3)(-5/3)x^3/3! + ....
=> 1 + x/3
Thus f(0.1) is approximated to 1 + 0.1/3 => 1.03

Answer 2

to locate the linear approximation, we ought to locate the tangent line to f(x) at x = one hundred twenty five. The slope of the tangent line to f(x) at x = one hundred twenty five, so we ought to locate f'(one hundred twenty five). f(x) has spinoff: f'(x) = (a million/3)x^(a million/3 - a million) = a million/[3x^(2/3)]. At x = one hundred twenty five: f'(one hundred twenty five) = a million/[3(one hundred twenty five)^(2/3)] = a million/seventy 5. Then, on account that f(x) passes by way of (one hundred twenty five, 5), so ought to the tangent line to f(x) at x = one hundred twenty five. for this reason, by way of element-slope style, the linear approximation is: y - 5 = (a million/seventy 5)(x - one hundred twenty five) ==> y = (a million/seventy 5)x + 10/3. --- to locate the approximation for (one hundred twenty five.a million)^(a million/3), keep in mind that the tangent line to f(x) at x = one hundred twenty five is a sturdy approximation for f(x) at x close to one hundred twenty five; it is: f(x) ? (a million/seventy 5)x + 10/3 for x ? one hundred twenty five. consequently: (one hundred twenty five.a million)^(a million/3) = f(one hundred twenty five.a million) ? (a million/seventy 5)(one hundred twenty five.a million) + 10/3 ? 5.00133. i'm hoping this enables!

Answer 3

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