sketch the graph of the function. identify any local extrema and points of inflections. f(x)=2x³ + 11/2x²…

sketch the graph of the function. identify any local extrema and points of inflections. f(x)=2x³ + 11/2x² - 7x + 4\n\nob. there is no minimum.\n\nwhat are the coordinates of the points of inflection? select the correct choice and, if necessary, fill in the answer box to complete your choice.\n\noa. \n(simplify your answer. type an ordered pair. type integers or simplified fractions. use a comma to separate answers as needed.)\n\nob. there are no inflection points.

sketch the graph of the function. identify any local extrema and points of inflections. f(x)=2x³ + 11/2x² - 7x + 4\n\nob. there is no minimum.\n\nwhat are the coordinates of the points of inflection? select the correct choice and, if necessary, fill in the answer box to complete your choice.\n\noa. \n(simplify your answer. type an ordered pair. type integers or simplified fractions. use a comma to separate answers as needed.)\n\nob. there are no inflection points.

Answer

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, we have $f(x)=2x^{3}+\frac{11}{2}x^{2}-7x + 4$. Then $f'(x)=6x^{2}+11x - 7$.

Step2: Solve for critical points

Set $f'(x)=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here $a = 6$, $b = 11$, $c=-7$. So $x=\frac{-11\pm\sqrt{11^{2}-4\times6\times(-7)}}{2\times6}=\frac{-11\pm\sqrt{121 + 168}}{12}=\frac{-11\pm\sqrt{289}}{12}=\frac{-11\pm17}{12}$. We get $x=\frac{-11 + 17}{12}=\frac{1}{2}$ and $x=\frac{-11-17}{12}=-\frac{7}{3}$.

Step3: Find the second - derivative

Differentiate $f'(x)=6x^{2}+11x - 7$ with respect to $x$. $f''(x)=12x + 11$.

Step4: Classify critical points

Evaluate $f''(x)$ at critical points. $f''(\frac{1}{2})=12\times\frac{1}{2}+11=6 + 11=17>0$, so $f(x)$ has a local minimum at $x = \frac{1}{2}$. $f(\frac{1}{2})=2\times(\frac{1}{2})^{3}+\frac{11}{2}\times(\frac{1}{2})^{2}-7\times\frac{1}{2}+4=2\times\frac{1}{8}+\frac{11}{2}\times\frac{1}{4}-\frac{7}{2}+4=\frac{1}{4}+\frac{11}{8}-\frac{7}{2}+4=\frac{2 + 11-28 + 32}{8}=\frac{17}{8}$. $f''(-\frac{7}{3})=12\times(-\frac{7}{3})+11=-28 + 11=-17<0$, so $f(x)$ has a local maximum at $x=-\frac{7}{3}$. $f(-\frac{7}{3})=2\times(-\frac{7}{3})^{3}+\frac{11}{2}\times(-\frac{7}{3})^{2}-7\times(-\frac{7}{3})+4=2\times(-\frac{343}{27})+\frac{11}{2}\times\frac{49}{9}+\frac{49}{3}+4=-\frac{686}{27}+\frac{539}{18}+\frac{49}{3}+4=\frac{-1372 + 1617+882 + 216}{54}=\frac{1343}{54}$.

Step5: Find points of inflection

Set $f''(x)=0$. $12x+11 = 0$, then $x=-\frac{11}{12}$. $f(-\frac{11}{12})=2\times(-\frac{11}{12})^{3}+\frac{11}{2}\times(-\frac{11}{12})^{2}-7\times(-\frac{11}{12})+4$. [ \begin{align*} f(-\frac{11}{12})&=2\times(-\frac{1331}{1728})+\frac{11}{2}\times\frac{121}{144}+\frac{77}{12}+4\ &=-\frac{1331}{864}+\frac{1331}{288}+\frac{77}{12}+4\ &=\frac{-1331 + 3993+4576+3456}{864}\ &=\frac{10694}{864}=\frac{5347}{432} \end{align*} ]

Answer:

The local minimum is at the point $(\frac{1}{2},\frac{17}{8})$, the local maximum is at the point $(-\frac{7}{3},\frac{1343}{54})$, and the point of inflection is at $(-\frac{11}{12},\frac{5347}{432})$