Use linear approximation to approximate 25.3‾‾‾‾√ as follows. Let f(x)=x√. The equation of the tangent line to f(x) at x=25 can be written

Question

Use linear approximation to approximate 25.3‾‾‾‾√ as follows. Let f(x)=x√. The equation of the tangent line to f(x) at x=25 can be written in the form y=mx+b. Compute m and b.

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Julia 1 week 2021-09-13T08:19:15+00:00 1 Answer 0

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    2021-09-13T08:20:56+00:00

    Answer:

    Approximation f(25.3)=5.03 (real value=5.0299)

    The approximation can be written as f(x)=0.1x+2.5

    Step-by-step explanation:

    We have to approximate f(25.3)=\sqrt{25.3} with a linear function.

    To approximate a function, we can use the Taylor series.

    f(x)=\sum_1^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

    The point a should be a point where the value of f(a) is known or easy to calculate.

    In this case, the appropiate value for a is a=25.

    Then we calculate the Taylor series with a number of terms needed to make a linear estimation.

    f(x)\approx f(a)+\frac{f'(a)}{1!}(x-a)

    The value of f'(a) needs the first derivate:

    f'(x)=\frac{1}{2\sqrt{x}}\\\\f'(a)=f'(25)=\frac{1}{2\sqrt{25}}=\frac{1}{2*5}=\frac{1}{10}

    Then

    f(x)\approx f(25)+\frac{1}{10}(x-25)=\sqrt{25} +\frac{1}{10}(x-25)\\\\f(x)\approx 5+\frac{1}{10}(x-25)

    We evaluate for x=25.3

    f(25.3)\approx 5+\frac{1}{10}(25.3-25)\\\\f(25.3)\approx 5+\frac{1}{10}(0.3)=5.03

    If we rearrange the approximation to be in the form mx+b we have:

    f(x)\approx 5+\frac{1}{10}(x-25)=5+0.1x-2.5\\\\f(x)\approx 0.1x+2.5

    Then, m=0.1 and b=2.5.

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