Let a be a real number such that the function | vedclass

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(D) Given $f(x) = ax^2 + 6x - 15$.$f'(x) = 2ax + 6$.Since $f(x)$ is increasing in $(-\infty, \frac{3}{4})$ and decreasing in $(\frac{3}{4}, \infty)$,t

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local maximum at $x = -\frac{3}{4}$

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(D) Given $f(x) = ax^2 + 6x - 15$.$f'(x) = 2ax + 6$.Since $f(x)$ is increasing in $(-\infty, \frac{3}{4})$ and decreasing in $(\frac{3}{4}, \infty)$,the critical point is $x = \frac{3}{4}$.At $x = \frac{3}{4}$,$f'(x) = 0$,so $2a(\frac{3}{4}) + 6 = 0 \Rightarrow \frac{3a}{2} = -6 \Rightarrow a = -4$.Now,consider $g(x) = ax^2 - 6x + 15$. Substituting $a = -4$,we get $g(x) = -4x^2 - 6x + 15$.$g'(x) = -8x - 6$.Setting $g'(x) = 0$,we get $-8x = 6 \Rightarrow x = -\frac{6}{8} = -\frac{3}{4}$.To check for local maximum or minimum,we use the second derivative test: $g''(x) = -8$.Since $g''(x) < 0$,the function $g(x)$ has a local maximum at $x = -\frac{3}{4}$.

Let $a$ be a real number such that the function $f(x) = ax^2 + 6x - 15, x \in R$ is increasing in $(-\infty, \frac{3}{4})$ and decreasing in $(\frac{3}{4}, \infty)$. Then the function $g(x) = ax^2 - 6x + 15, x \in R$ has a:

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