Question

maximum and minimum value of the function Y is equals to x cube minus 3 X square + 6​

Answer 1

It is based on Application of derivatives concepts .

Given, y = x³ - 3x² + 6

Now, differentiate with respect to x

dy/dx = 3x² - 6x + 0

∴ at dy/dx = 0 = 3x² - 6x

x(x - 2) = 0

⇒ x = 0, 2

For maximum d²y/dx² < 0 at x = 0 or 2

For minimum d²y/dx² > 0 at x = 0 or 2

Let's check each of them ,

differentiate dy/dx with respect to x

d²y/dx² = 6x - 6

Put x = 0 , d²y/dx² = 0 - 6 < 0

Hence, at x = 0 , y will be maximum

So, maximum value of y = (0)³ - 3(0)² + 6 = 6

Put x = 2 , d²y/dx² = 6 × 2 - 6 > 0

Hence at x = 2 , y will be minimum

So, minimum value of y = (2)³ - 3(2)² + 6

= 8 - 12 + 6 = 2