Let f(x) = ax^2 - b|x|,where a and b are cons | vedclass

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(A) Given $f(x) = ax^2 - b|x|$.For $x \geq 0$,$f(x) = ax^2 - bx$. The derivative is $f'(x) = 2ax - b$. At $x=0$,$f'(0) = -b$.For $x < 0$,$f(x) = ax^2

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a maxima whenever $a > 0, b > 0$

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(A) Given $f(x) = ax^2 - b|x|$.For $x \geq 0$,$f(x) = ax^2 - bx$. The derivative is $f'(x) = 2ax - b$. At $x=0$,$f'(0) = -b$.For $x Case $1$: If $a > 0$ and $b > 0$,then for $x$ slightly greater than $0$,$f(x) \approx -bx Case $2$: If $a > 0$ and $b  0$. Then $f(x) = ax^2 + k|x|$. Since $a > 0$ and $k > 0$,$f(x) > 0$ for all $x 
eq 0$ and $f(0) = 0$. Thus,$f(x)$ has a local minima at $x = 0$.Therefore,the correct statement is that $f(x)$ has a maxima whenever $a > 0, b > 0$.

Let $f(x) = ax^2 - b|x|$,where $a$ and $b$ are constants. Then at $x = 0$,$f(x)$ has

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