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How do you find the linearization at a=81 of #f(x) = x^(3"/"4)#?

Answer 1

Emily Ammerman

#L = f(a)+f'(a)(x-a)# (This is an equation of the tangent line at #(a,f(a))#)

#f(x) = x^(3/4)#, #a = 81#

#f(81) = (root(4)81)^3 = 3^3 = 27#

#f'(x) = 3/(4x^(1/4))#, so #f'(81) = 3/(4root(4)81) = 3/4*1/4#

#L = 27+1/4(x-27)#

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Answer 2

Paisley Johnson

To find the linearization of (f(x) = x^{3/4}) at (a = 81), follow these steps:

Find the first derivative of (f(x)). [f'(x) = \frac{3}{4}x^{-1/4}]
Evaluate (f'(a)) at (a = 81). [f'(81) = \frac{3}{4}(81)^{-1/4} = \frac{3}{4 \times 3} = \frac{1}{4}]
Find the value of (f(a)) at (a = 81). [f(81) = 81^{3/4} = (3)^3 = 27]
Use the formula for linearization: [L(x) = f(a) + f'(a)(x - a)]
Substitute (a = 81), (f(a) = 27), (f'(a) = \frac{1}{4}) into the linearization formula: [L(x) = 27 + \frac{1}{4}(x - 81)]

Thus, the linearization of (f(x)) at (a = 81) is: [L(x) = 27 + \frac{1}{4}(x - 81)]

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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.