"Let 'a' be a real number such that the function f(x) = ax2 + 6x $-$ 15, x $\in$ R is increasing in $\left( { - \infty ,{3 \over 4}} \right)$ and decreasing in $\left( {{3 \over 4},\infty } \right)$. Then the function g(x) = ax2 $-$ 6x + 15, x$\in$R has a :"

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Accepted Answer

"${{ - B} \over {2A}} = {3 \over 4}$

$\Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}$

$\Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4$

$\therefore$ $g(x) = 4{x^2} - 6x + 15$

Local max. at $x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$

$= {{ - 3} \over 4}$"

Let 'a' be a real number such that the function f(x) = ax² + 6x $-$ 15, x $\in$ R is increasing in $\left( { - \infty ,{3 \over 4}} \right)$ and decreasing in $\left( {{3 \over 4},\infty } \right)$. Then the function g(x) = ax² $-$ 6x + 15, x$\in$R has a :

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Let 'a' be a real number such that the function f(x) = ax2 + 6x $-$ 15, x $\in$ R is increasing in $\left( { - \infty ,{3 \over 4}} \right)$ and decreasing in $\left( {{3 \over 4},\infty } \right)$. Then the function g(x) = ax2 $-$ 6x + 15, x$\in$R has a :

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Let 'a' be a real number such that the function f(x) = ax² + 6x $-$ 15, x $\in$ R is increasing in $\left( { - \infty ,{3 \over 4}} \right)$ and decreasing in $\left( {{3 \over 4},\infty } \right)$. Then the function g(x) = ax² $-$ 6x + 15, x$\in$R has a :