Prove that the function given by f(x)=x^3-3x^2+3x-100 is increasing in R.
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f(x) = x³ - 3x² + 3x - 100
differentiate with respect to x,
f'(x) = d(x³ - 3x² + 3x - 100)/dx
= d(x³)/dx - 3d(x²)/dx + 3dx/dx - d(100)/dx
= 3x² - 6x + 3
= 3(x² - 2x + 1)
= 3(x - 1)²
what you see ?, f'(x) = 3(x - 1)² > 0 for all real value of x.
e.g,. f'(x) > 0 for all real value of x.
hence, f(x) is increasing for all real value of x
differentiate with respect to x,
f'(x) = d(x³ - 3x² + 3x - 100)/dx
= d(x³)/dx - 3d(x²)/dx + 3dx/dx - d(100)/dx
= 3x² - 6x + 3
= 3(x² - 2x + 1)
= 3(x - 1)²
what you see ?, f'(x) = 3(x - 1)² > 0 for all real value of x.
e.g,. f'(x) > 0 for all real value of x.
hence, f(x) is increasing for all real value of x
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