I am to graph the function but i cant find where f''(x) equals zero

**Alioyo** · October 27th 2012, 10:06 AM

f(x)= x^4 - 4x^3 - 8x^2 + 48x
f'(x)= 4 (x-2)(x+2)(x-3)
f''(x)= 4(3x^2 -6x- 4)

my problem is i am unable to get the concavity.
what would be my intervals?

**psilver1** · October 27th 2012, 11:26 AM

Have you tried factoring the 3x^2-6x-4 and then setting those to 0

**Soroban** · October 27th 2012, 01:27 PM

Hello, Alioyo!

$f(x)\:=\: x^4 - 4x^3 - 8x^2 + 48x$
$f'(x)\:=\: 4x^3-12x^2 - 16x + 48 \:=\: 4 (x-2)(x+2)(x-3)$
$f''(x)\:=\: 4(3x^2 -6x- 4)$

My problem is i am unable to get the concavity. .Why not?
What would be my intervals?

You're taking Calculus and you don't know how to solve a quadratic equation?

You have: . $3x^2-6x-4 \:=\:0$

Quadratic Formula: . $x \:=\:\frac{6 \pm\sqrt{36+48}}{6} \:=\:\frac{6\pm\sqrt{84}}{6} \:=\:\frac{6\pm2\sqrt{21}}{6} \:=\:\frac{3\pm\sqrt{21}}{3}$

The intervals are: . $\left(\text{-}\infty,\:\tfrac {3 -\sqrt{21}}{3} \right),\;\left(\tfrac{3-\sqrt{21}}{3},\:\tfrac{3+\sqrt{21}}{3}\right),\; \left(\tfrac{3+\sqrt{21}}{3},\:\infty\right)$

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