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(a) Sove 10)21 The solution to 0-0ax Type an integer or a fraction Use a comma t

ID: 3116303 • Letter: #

Question

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Explanation / Answer

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Given

f(x) = -x^2 + 1

g(x) = 6x + 1

a) f(x) = 0

=> -x^2 + 1 = 0

x^2 = 1

x = +/- 1

b) g(x) = 0

6x + 1 = 0

x = -1/6

c) f(x) = g(x)

-x^2 + 1 = 6x + 1

x^ + 6x = 0

x ( x + 6) = 0

x = 0, -6

d) f(x) > 0

-x^2 + 1 > 0

x^2 - 1 < 0

(x - 1) (x+ 1) < 0

Hence the value of f(x) > 0 when -1<x <1

Solution