If the function y = a log x + bx^2 + x has ex | vedclass

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(D) Given $y = a \log x + bx^2 + x$. Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extreme values at $x = 1$ an

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$\left( -\frac{2}{3}, -\frac{1}{6} \right)$

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(D) Given $y = a \log x + bx^2 + x$. Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extreme values at $x = 1$ and $x = 2$,the derivative must be zero at these points. At $x = 1$: $\frac{a}{1} + 2b(1) + 1 = 0 \implies a + 2b = -1$ (Equation $1$). At $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b = -1 \implies a + 8b = -2$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - (-1) \implies 6b = -1 \implies b = -\frac{1}{6}$. Substitute $b = -\frac{1}{6}$ into Equation $1$: $a + 2(-\frac{1}{6}) = -1 \implies a - \frac{1}{3} = -1 \implies a = -1 + \frac{1}{3} = -\frac{2}{3}$. Thus,$(a, b) = \left( -\frac{2}{3}, -\frac{1}{6} \right)$.

If the function $y = a \log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$,then $(a, b) = \dots$

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