# What is the local linearization of #y = (7+5x^2)^(-1/2)# at a=0?

It is

Writing the equation of the tangent line in point-slope form, we get

graph{y=(7+5x^2)^(-1/2) [-3.467, 3.465, -1.35, 2.117]}

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The local linearization of ( y = (7 + 5x^2)^{-1/2} ) at ( a = 0 ) is given by the equation: [ L(x) = y'(a)(x - a) + y(a) ] where ( y'(a) ) represents the derivative of ( y ) with respect to ( x ) evaluated at ( x = a ). The derivative of ( y ) with respect to ( x ) is: [ y'(x) = \frac{-5x}{2(7 + 5x^2)^{3/2}} ] Evaluating ( y'(0) ), we get: [ y'(0) = \frac{0}{2(7 + 5(0)^2)^{3/2}} = 0 ] Next, ( y(0) ) can be found by substituting ( x = 0 ) into the original function: [ y(0) = (7 + 5(0)^2)^{-1/2} = (7)^{-1/2} = \frac{1}{\sqrt{7}} ] Thus, the local linearization of ( y = (7 + 5x^2)^{-1/2} ) at ( a = 0 ) is: [ L(x) = 0 \cdot (x - 0) + \frac{1}{\sqrt{7}} = \frac{1}{\sqrt{7}} ]

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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.

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.

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.

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- At noon, aircraft carrier Alpha is 100 kilometers due east of destroyer Beta. Alpha is sailing due west at 12 kilometers per hour. Beta is sailing south at 10 kilometers per hour. In how many hours will the distance between the ships be at a minimum?

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