Question

f(x)= sqrt(x^4-16x^2)
a)Find the domain of f.
b) Find f '(x)
c) Find the slope of the line normal to the graph of f at x = 5

Answer 1

a)x^4 - 16x^2 >= 0
x^2 (x^2 - 16) >= 0
x^2 (x + 4)(x - 4) >= 0

since x^2 is always >= 0, this will be positive either when both of the other factors are negative (x < -4) or positive (x > 4)

Answer 2

b)f ' = (4x^3 - 32x) / [2sqrt(x^4 - 16x^2)] or f ' (x) = (2x^3 - 16x) / sqrt(x^4 - 16x^2)

Answer 3

c) f ' (5) = (2*125 - 16(5)) / sqrt(625 - 16(25))
f ' (5) = (250 - 80) / sqrt(225) = 170 / 15 = 34 /3
f(5) = sqrt(625 - 16(25)) = sqrt(225) = 15
point of tangency, and point for normal: (5 , 15)
normal line: 15 = -3/34 (5) + b
b = 15 + 15/34 = 525/34
normal line: y = (-3/34)x + 525/34

Answer 4

thank you