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

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=2x^{3}+\frac{11}{2}x^{2}-7x + 4$ using the power rule. $f'(x)=6x^{2}+11x - 7$.

Step2: Solve for critical points

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

Step4: Classify critical points

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

Answer:

What are the coordinates of the relative maxima (also called local maxima)? A. $(-\frac{7}{3},\frac{1343}{54})$ What are the coordinates of the relative minima (also called local minima)? A. $(\frac{1}{2},\frac{17}{8})$