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How do you use a linear approximation or differentials to estimate #(2.001)^5#?

Answer 1

William Abdalla

See below

We know that if #f(x)# is sufficiently differentiable then

#f(a+delta)approxf(a)+f^((1))(a)/(1!)delta + f^((2))(a)/(2!) delta^2+...+f^((n))(a)/(n!)delta^n + cdots#

Here #f^((i))# indicates the #i#-th derivative

Now making #a = 2# and #delta = 0.001#

#(2.001)^5 approx 2^5 + 5/(1!)cdot 2^4 cdot 0.001=32.08#

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Answer 2

Isaiah Allen

To estimate ( (2.001)^5 ) using linear approximation or differentials, you can use the formula:

[ f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x ]

Where:

( f(x) = x^5 ) is the function you want to approximate.
( x = 2 ) is the point you're approximating around.
( \Delta x = 0.001 ) is the small change in ( x ).

First, find the derivative of ( f(x) = x^5 ): [ f'(x) = 5x^4 ]

Then evaluate ( f'(x) ) at ( x = 2 ): [ f'(2) = 5(2)^4 = 80 ]

Now, apply the formula: [ f(2.001) \approx f(2) + f'(2) \cdot 0.001 ] [ \approx 2^5 + 80 \cdot 0.001 ] [ \approx 32 + 0.08 ] [ \approx 32.08 ]

So, ( (2.001)^5 ) is approximately 32.08.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.