y=x^3-a x^2+48 x+7 is an increasing function | vedclass

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Question

(B) Given the function $y=x^3-a x^2+48 x+7$. For the function to be an increasing function for all real values of $x$,its derivative must be non-negat

Vedclass · Accepted Answer

$(-12, 12)$

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(B) Given the function $y=x^3-a x^2+48 x+7$. For the function to be an increasing function for all real values of $x$,its derivative must be non-negative,i.e.,$\frac{dy}{dx} \geq 0$ for all $x \in \mathbb{R}$. Differentiating with respect to $x$: $\frac{dy}{dx} = 3x^2 - 2ax + 48$. Since $3x^2 - 2ax + 48 \geq 0$ for all $x$,the quadratic expression must have a discriminant $D \leq 0$. Here,$A=3, B=-2a, C=48$. $D = B^2 - 4AC = (-2a)^2 - 4(3)(48) = 4a^2 - 576$. Setting $D \leq 0$: $4a^2 - 576 \leq 0$ $4(a^2 - 144) \leq 0$ $a^2 - 144 \leq 0$ $(a-12)(a+12) \leq 0$. Thus,$a \in [-12, 12]$. Given the options,the interval is $(-12, 12)$.

$y=x^3-a x^2+48 x+7$ is an increasing function for all real values of $x$,then $a$ lies in the interval

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