
29 Jun, 2021 03:15 PM
how would i solve this?
I'm having a lot of trouble understanding math, some help would be appreciative. the question is to determine zeros from following equations: a) g(x) = 2x^2 - x - 6 b) h(x) = 2^x - 1 c) j(x) = sin (x - 45degrees), where 0degrees < or = x < or = 360degrees d) k(x) = 2 cosx, where -360degrees < or = x < or = 0degrees
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a) g(x) = 2x^2 - x - 6
Ans:
The zeroes of g(x) is given when,
2x^2 - x - 6 = 0
2x^2 - 4x + 3x - 6 = 0
2x(x - 2) + 3(x - 2) = 0
(2x + 3)(x - 2) = 0
2x + 3 = 0
2x = -3
x = -3/2 or
x - 2 = 0
x = 2
The zeroes are when x = -3/2 and 2
b) h(x) = 2^x - 1
Ans:
2x^2 - 1 = 0
2x^2 = 1
x^2 = 1/2
x = ±√(1/2)
The zeroes are when x = -√(1/2) and +√(1/2)
c) j(x) = sin (x - 45degrees), where 0degrees < or = x < or = 360degrees
Ans:
When j(x) = 0
Sin (x - 45) = 0
Sin^-1 (0) = x - 45
0 = x - 45
45 = x
Recall sin a = sin (180 - a)
x - 45 = 180
x = 180 + 45
x = 225
Thus, the zeroes are when x = 45 degrees and 225 degrees
d) k(x) = 2 cosx, where -360degrees < or = x < or = 0degrees
Ans:
When k(x) = 0,
2cosx = 0
Cosx = 0/2 = 0
Cosx = 0
Cos^-1 0 = x
x = 90