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Question Number 157382 by cortano last updated on 22/Oct/21

Answered by mr W last updated on 23/Oct/21

$$g\left(x\right)=x+\frac{1}{x^2}$$$$g^{\prime }\left(x\right)=1-\frac{2}{x^3}=0\Rightarrow x=\sqrt[3]{2}\in [1,2]$$$$g{\left(x\right)}_{min}=\frac{3\sqrt[3]{2}}{2}=k$$$$f\left(x\right)=x^2+ax+b={(x+\frac{a}{2})}^2+(b-\frac{a^2}{4})$$$$f{\left(x\right)}_{min}=b-\frac{a^2}{4}=k=\frac{3\sqrt[3]{2}}{2}$$$$f{\left(x\right)}_{max}=f\left(2\right)=4+2a+b=4+2a+\frac{a^2}{4}+\frac{3\sqrt[3]{2}}{2}$$

Commented by cortano last updated on 24/Oct/21

$$f{\left(x\right)}_{min}=b-\frac{a^2}{4}=\frac{3\sqrt[3]{2}}{2};whena=-4$$$$sof\left(x\right)=x^2-4x+b$$$$\{\begin{array}{l}f\left(1\right)=1-4+4+\frac{3\sqrt[3]{2}}{2}=1+\frac{3\sqrt[3]{2}}{2}\\ f\left(2\right)=4-8+4+\frac{3\sqrt[3]{2}}{2}=\frac{3\sqrt[3]{2}}{2}\end{array}$$$$maxf\left(x\right)=f\left(1\right)=1+\frac{3\sqrt[3]{2}}{2}$$

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