Question

Find the maximum and minimum

y=x^3-6x^2+9x+15

Using differentiation.

Answer 1

$\bf{\underline{\underline{Given}}}$

$\mathtt{y = x^{3} - 6x^{2} + 9x + 15}$

$\bf{\underline{\underline{To \:find}}}$

$\mathsf {Mamimum \: and \: Minimum}$

$\bf{\underline{\underline{Solution}}}$

First, we have to find the derivatives of y using differentiation

$\mathtt{ y = x^{3} - 6x^{2} + 9x + 15}$

For Maximum and Minimum

$\mathtt{\implies\: \frac{dy}{dx} = 0}$

$\mathtt{\implies\: \frac{d}{dx}( x^{3} - 6x^{2} + 9x + 15) = 0}$

$\mathtt{\implies\: 3x^{2} - 12x + 9 = 0}$

Solving the eqn, we get:

$\mathtt{\implies\: 3x^{2} - 9x - 3x + 9 = 0}$

$\mathtt{\implies\: 3x(x - 3) - 3(x - 3) = 0}$

$\mathtt{\implies\: (3x - 3) (x - 3) = 0}$

$\mathtt{\implies\: x = 1 \: \: or \ \: x = 3}$

We know that

• (dy/dx) is negative at maximum
• (dy/dx) is positive at minimum

Here, we have both positive values. So we can further differentiate dy/dx

$\mathtt{\implies\: \frac{dy}{dx} = 3x^{2} - 12x + 9}$

$\mathtt{\implies\: \frac{d \: (\frac{dy}{dx})}{dx}}$

$\mathtt{\implies\: \frac{d}{dx} (3x^{2} - 12x + 9)}$

$\mathtt{\implies\: 6x - 12}$

Putting values of x

☞ 6(1) - 12 = - 6 (Maximum)

☞ 6(3) - 12 = 6 (Minimum)

∴ y is maximum at (-6) and minimum at 6