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secondary 4 | A Maths
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Date Posted: 1 year ago
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y = (2x–3)/(3x–5)
dy/dx = [(2)(3x–5)–(3)(2x–3)]/(3x–5)² = (6x–10–6x+9)/(3x–5)² = –1/(3x–5)²
Since –1/(3x–5)² ≠ 0, therefore dy/dx ≠ 0 and curve has no turning point
dy/dx = [(2)(3x–5)–(3)(2x–3)]/(3x–5)² = (6x–10–6x+9)/(3x–5)² = –1/(3x–5)²
Since –1/(3x–5)² ≠ 0, therefore dy/dx ≠ 0 and curve has no turning point
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