Cal 1
posted by Dee .
Find the linearization L(x) of the function at a.
f(x) = x^4/5, a = 32

Assuming you mean x^(4/5), else x^4 is huge at x=32,
f(32) = 16
f' = 4/5 x^(1/5)
f'(32) = 4/5 * 1/2 = 2/5
So, now you have a point (32,16) and a slope (2/5). What's the line?
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