# 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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