If a cubic function f(x)=a x^3+b x^2-frac{18} | vedclass

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(B) Given the function $f(x)=a x^3+b x^2-\frac{18}{5} x+\frac{19}{10}$.Taking the derivative,$f^{\prime}(x)=3 a x^2+2 b x-\frac{18}{5}$.Since $f(x)$ h

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$\frac{-6}{5}$

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(B) Given the function $f(x)=a x^3+b x^2-\frac{18}{5} x+\frac{19}{10}$.Taking the derivative,$f^{\prime}(x)=3 a x^2+2 b x-\frac{18}{5}$.Since $f(x)$ has a maximum at $x=-3$,$f^{\prime}(-3)=0$:$3 a(-3)^2+2 b(-3)-\frac{18}{5}=0 \Rightarrow 27 a-6 b=\frac{18}{5} \Rightarrow 9 a-2 b=\frac{6}{5} \cdots (1)$.Since $f(x)$ has a minimum at $x=2$,$f^{\prime}(2)=0$:$3 a(2)^2+2 b(2)-\frac{18}{5}=0 \Rightarrow 12 a+4 b=\frac{18}{5} \Rightarrow 6 a+2 b=\frac{9}{5} \cdots (2)$.Adding $(1)$ and $(2)$:$(9 a-2 b)+(6 a+2 b)=\frac{6}{5}+\frac{9}{5} \Rightarrow 15 a=\frac{15}{5} \Rightarrow 15 a=3 \Rightarrow a=\frac{1}{5}$.Substituting $a=\frac{1}{5}$ into $(2)$:$6(\frac{1}{5})+2 b=\frac{9}{5} \Rightarrow 2 b=\frac{9}{5}-\frac{6}{5}=\frac{3}{5} \Rightarrow b=\frac{3}{10}$.Thus,$f(x)=\frac{1}{5} x^3+\frac{3}{10} x^2-\frac{18}{5} x+\frac{19}{10}$.Calculating $f(1)$:$f(1)=\frac{1}{5}(1)^3+\frac{3}{10}(1)^2-\frac{18}{5}(1)+\frac{19}{10} = \frac{2}{10}+\frac{3}{10}-\frac{36}{10}+\frac{19}{10} = \frac{2+3-36+19}{10} = \frac{-12}{10} = \frac{-6}{5}$.Therefore,$f(1)=\frac{-6}{5}$.

If a cubic function $f(x)=a x^3+b x^2-\frac{18}{5} x+\frac{19}{10}$ has a maximum value of $10$ at $x=-3$ and a minimum value of $\frac{-5}{2}$ at $x=2$,then $f(1)=$

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