Problems from Concavity and convexity

Development:

• We determine the zeros or solutions of the function:

$f(x)=(x-2{)}^2(x+1)=0\Rightarrow x=2;x=-1$

• We determine the relative extrema and the inflection points:

$f^{\prime }(x)=2(x-2)(x+1)+(x-2{)}^2=(x-2)(2(x+1)+x-2)=3x(x-2)$

We equate the first derivative to zero to find maxima and minima:

$f^{\prime }(x)=3x(x-2)=0\Rightarrow x=2;x=0.(f(2)=0;f(0)=4)$

We look at the second derivative in each of the values and look at its sign:

$f^{″}(x)=3(x-2)+3x=6x-6$

$f^{″}(2)=5>0\Rightarrow Min$

$f^{″}(0)=-6<0\Rightarrow Max$

Therefore, the relative extrema are

$(2,0)Min$; $(0,4)Max$

Let's see the inflection point:

$f^{″}(x)=6x-6=0\Rightarrow x=1(f(1)=2)$

Therefore there is an inflection point at $(1,2)$.

• To study the increasing / decreasing intervals we must analyze the derivative function $f^{\prime }(x)$, $f^{\prime }(x)=3x(x-2)$.

We have seen the roots, which give us the intervals. Let's see every interval and the value of the derivative in it.

$(-\mathrm{\infty },0)f^{\prime }(-10)>0\Rightarrow$ Increasing

$(0,2)f^{\prime }(1)<0\Rightarrow$ Decreasing

$(2,\mathrm{\infty })f^{\prime }(10)>0\Rightarrow$ Increasing

To study the concavity we must look at the second derivative in every interval (now the intervals are given by the zeros of the second derivative, not of the first one as earlier): $f^{″}(x)=6x-6$.

The intervals are:

$(-\mathrm{\infty },1)f^{″}(-10)<0\Rightarrow$ Concave

$(1,\mathrm{\infty })f^{″}(10)>0\Rightarrow$ Convex

Solution:

• $x=2;x=-1$
• $(2,0)Min$; $(0,4)Max$; Inflection point at $(1,2)$.
• Increase/decrease intervals:

$(-\mathrm{\infty },0)f^{\prime }(-10)>0\Rightarrow$ Increasing

$(0,2)f^{\prime }(1)<0\Rightarrow$ Decreasing

$(2,\mathrm{\infty })f^{\prime }(10)>0\Rightarrow$ Increasing

Convexity / concavity intervals:

$(-\mathrm{\infty },1)f^{″}(-10)<0\Rightarrow$ Concave

$(1,\mathrm{\infty })f^{″}(10)>0\Rightarrow$ Convex

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