Jun Heng's answer to Ray's Secondary 4 A Maths Singapore question.
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3i)
y = x³+3x
dy/dx = 3x²+3
3ii)
Since x² > 0 for all values of x, therefore 3x²+3 > 0, hence dy/dx > 0, therefore y increases as x increases
4)
y = 2x/(x+1)
dy/dx = [2(x+1)–(2x)(1)]/(x+1)² = (2x+2–2x)/(x+1)² = 2/(x+1)²
Since (x+1)² > 0 for all values of x, therefore 2/(x+1)² > 0, hence dy/dx > 0, therefore y increases as x increases
y = x³+3x
dy/dx = 3x²+3
3ii)
Since x² > 0 for all values of x, therefore 3x²+3 > 0, hence dy/dx > 0, therefore y increases as x increases
4)
y = 2x/(x+1)
dy/dx = [2(x+1)–(2x)(1)]/(x+1)² = (2x+2–2x)/(x+1)² = 2/(x+1)²
Since (x+1)² > 0 for all values of x, therefore 2/(x+1)² > 0, hence dy/dx > 0, therefore y increases as x increases
Date Posted:
1 year ago
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