If $f(x)=x^5-5 x^4+5 x^3-10$ has its local maxima and minima at $x=a$ and $x=b$ respectively, then $2 a+b$ is equal to

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Aicharya · Accepted Answer

We have, $ \begin{aligned} f(x) & =x^5-5 x^4+5 x^3-10 \ f^{\prime}(x) & =5 x^4-20 x^3+15 x^2 \end{aligned} $ For maxima or minima put $f^{\prime}(x)=0$ $ \begin{aligned} 	herefore \quad 5 x^4-20 x^3+15 x^2 & =0 \ \Rightarrow \quad 5 x^2\left(x^2-4 x+3ight) & =0 \ \Rightarrow \quad 5 x^2(x-3)(x-1) & =0 \ x & =0,1,3 \ f^{\prime \prime}(x) & =20 x^3-60 x^2+30 x \ f^{\prime \prime}(x) & =10 x\left(2 x^2-6 x+3ight) \ f^{\prime \prime}(0) & =0 \ f^{\prime \prime}(1) & =10(2-6+3) 0 \end{aligned} $ $	herefore \quad$ Local maxima at $x=1$ and local minima at $x=3$ Here, $a=1, b=3$ $ 	herefore \quad 2 a+b=2+3=5 $

If $f(x)=x^5-5 x^4+5 x^3-10$ has its local maxima and minima at $x=a$ and $x=b$ respectively, then $2 a+b$…

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