1. Differentiation Problem!!

Question:
Find the coordinates of the stationary points on the graph of the following functions, and find weather these points are maxima or minima.
$\displaystyle (a) 2x^3+3x^2-72x+5$

Attempt:
$\displaystyle f(x) = 2x^3+3x^2-72x+5$
$\displaystyle f'(x) = 6x^2+6x-72$
$\displaystyle f'(x) = 0$

$\displaystyle 6x^2+6x-72 = 0$
$\displaystyle a=1, b=1, c=-72$
$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle x = \frac{-1 \pm \sqrt{1^2-4\times1\times-72}}{2\times1}$

$\displaystyle x = \frac{-1 \pm 17}{2}$

$\displaystyle x = \frac{-1 + 17}{2}$ $\displaystyle x=8$

$\displaystyle x = \frac{-1 - 17}{2}$ $\displaystyle x=-9$

$\displaystyle f'(2) = 6x^2+6x-72$
$\displaystyle f'(2) = 6(2)^2+6(2)-72$
$\displaystyle f'(2) = -36$

$\displaystyle f'(12) = 6x^2+6x-72$
$\displaystyle f'(12) = 6(12)^2+6(12)-72$
$\displaystyle f'(12) = 864$

$\displaystyle x \leq 8$ ; minima

Where did I go wrong?

2. It would appear your error is in the use of the quadratic formula.

Your quadratic is $\displaystyle f'(x)=x^{2}+x-12=0$

3. Question:
Find the coordinates of the stationary points on the graph of the following functions, and find weather these points are maxima or minima.
$\displaystyle (a) 2x^3+3x^2-72x+5$

Attempt:
$\displaystyle f(x) = 2x^3+3x^2-72x+5$
$\displaystyle f'(x) = 6x^2+6x-72$
$\displaystyle f'(x) = 0$

$\displaystyle 6x^2+6x-72 = 0$
$\displaystyle a=1, b=1, c=-12$
$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle x = \frac{-1 \pm \sqrt{1^2-4\times1\times-12}}{2\times1}$

$\displaystyle x = \frac{-1 \pm 7}{2}$

$\displaystyle x = \frac{-1 + 7}{2}$ $\displaystyle x=3$

$\displaystyle x = \frac{-1 - 17}{2}$ $\displaystyle x=-4$

$\displaystyle f'(3) = 6x^2+6x-72$
$\displaystyle f'(3) = 6(3)^2+6(3)-72$
$\displaystyle f'(3) = 0$

$\displaystyle f'(-4) = 6x^2+6x-72$
$\displaystyle f'(-4) = 6(-4)^2+6(-4)-72$
$\displaystyle f'(-4) = 0$

I don't know what to do next... answer in the text book is $\displaystyle (-4,213)$$\displaystyle ,maximum; \displaystyle (3,-130)$$\displaystyle ,minimum$

$\displaystyle x^{2}+x-12=(x-3)(x+4)$