Given the function f(x)=112x4+16x3−3x2−2x+1 how do you find any points of inflection and determine where the curve is concave up or down?

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Alexzander Evans · Accepted Answer

Step 1If f is twice differentiable on an open set A, the inflections are the points where f&#039;&#039; changes sing. This is a polynomial so it&#039;s obvious that it&#039;s smooth (i.e. infinitively differentiable) on R, so here, the inflection points are the points where f0 and exist n&amp;gt;1 such that f(2n+1)≠0.So fx2+x−6 and its roots are −1±1+242={2,−3}Step 2Now we notice that the third derivative: 2x+1 has zero only in 1, so it&#039;s not zero in the candidate inflection points, and this tells us that -3 is a falling point and 2 is a rising point(this does make sense since we know how the graphic of a quartic polynomial behave)Now, if the function is twice differentiable, the concavity is determined by the sign of its second derivative:f0∈U={x&amp;lt;−3 or x&amp;gt;2}f0∈D={−3&amp;lt;x&amp;lt;2}So in U the curve is concave up, and in D the curve is concave down

Given the function \(\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{{12}}}}{x}^{{{4}}}+{\frac{{{1}}}{{{6}}}}{x}^{{{3}}}-{3}{x}^{{{2}}}-{2}{x}+{1}\) how do you

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