Use linear approximation to estimate 15^.25

Let f(x) = sqrt(x). Use linear approximation about x = 25, since 24 is close to 25, and since it is easy to take the square root of 25. Note that f ' (x) = 1/[2sqrt(x)] from the power rule. About x = 25, f(x) is approximately f(25) + f ' (25)(x - 25). So sqrt(24) = f(24) is approximately f(25) + f ' (25)(24 - 25) = sqrt(25) + {1/[2sqrt(25)]}(-1) = 5 - 1/10 = 4.9 .

A linear approximation of f(x) in a neighborhood of a 


f(x) ~ f(a) + f'(a) (x-a) 


f(x) = x^1/4 

2^4 = 16 i.e. a = 16 

f(15) ~ f(16) + f'(16) (15-16) 

f'(x) = 1/4 x^(-3/4) 

f(15) ~ 2 + 1/4 (16)^-3/4 (-1) 

~ 2 - 1/(32) = 1.96875

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Answer: v= lwhstep by step explanation: length x width x heightl = 12kmw = 6kmh = 6km12 x 6 = 7272 x 6 = 432therefore your answer would be c) 432 km^3