If y = a ln |x + 1| + b(x + 1)^2 + x has an e | vedclass

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(A) Given the function $y = a \ln |x + 1| + b(x + 1)^2 + x$. Since the function has an extremum at $x = 0$,the value of the function at $x = 0$ is $4$

Vedclass · Accepted Answer

$(-9, 4)$

Vedclass · Answer

(A) Given the function $y = a \ln |x + 1| + b(x + 1)^2 + x$. Since the function has an extremum at $x = 0$,the value of the function at $x = 0$ is $4$. Substituting $x = 0$ and $y = 4$ into the equation: $4 = a \ln |0 + 1| + b(0 + 1)^2 + 0$ $4 = a \ln(1) + b(1)^2 + 0$ Since $\ln(1) = 0$,we get $4 = 0 + b + 0$,which implies $b = 4$. Now,find the derivative $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{a}{x + 1} + 2b(x + 1) + 1$. At an extremum point,the derivative is zero. Thus,at $x = 0$,$\frac{dy}{dx} = 0$: $\frac{a}{0 + 1} + 2b(0 + 1) + 1 = 0$ $a + 2b + 1 = 0$. Substitute $b = 4$ into the equation: $a + 2(4) + 1 = 0$ $a + 8 + 1 = 0$ $a + 9 = 0 \Rightarrow a = -9$. Therefore,$(a, b) = (-9, 4)$.

If $y = a \ln |x + 1| + b(x + 1)^2 + x$ has an extremum value of $4$ at $x = 0$,then the value of $(a, b)$ is:

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