Is $f (x) = (x - 3) (x + 2) {(x - 4)}^{2}$ $f(x)=(x-3)(x+2)(x-4)^2$ concave or convex at $x = - 1$ $x=-1$ ?

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Answer 1 · 2017-03-07T14:19:27

The function is convex at $x = - 1$ $x=-1$

We need

$(uvw)'=u'vw+v'uw+w'uv$

We must calculate the first and second derivatives

$f(x)=(x-3)(x+2)(x-4)^2$

$f'(x)=(x+2)(x-4)^2+(x-3)(x-4)^2+2(x-3)(x+2)(x-4)$

$=(x-4){(x+2)(x-4)+(x-3)(x-4)+2(x-3)(x+2)}$

$=(x-4){x^2-2x-8+x^2-7x+12+2x^2-2x-12}$

$=(x-4)(4x^2-11x-8)$

$f''(x)=(4x^2-11x-8)+(x-4)(8x-11)$

$=4x^2-11x-8+8x^2-43x+44$

$=12x^2-54x+36$

Therefore,

$f''(-1)=12+54+36=102$

As, $f''(-1)>0$ , we conclude that the function is convex

Is f(x)=(x-3)(x+2)(x-4)^2f(x)=(x−3)(x+2)(x−4)2f(x)=(x-3)(x+2)(x-4)^2 concave or convex at x=-1x=−1x=-1?