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

Given,

$$f(x)=ax^2+6x-15$$

Then,

$$f'(x)=2ax+6$$

Since

$$f(x)$$ is increasing before

$$x=\frac34$$

and decreasing after it, the point $$x=\frac34$$ is a point of local maximum.

Hence,

$$f'\left(\frac34\right)=0$$

Therefore,

$$2a\left(\frac34\right)+6=0$$

$$\frac{3a}{2}=-6$$

$$a=-4$$

Now,

$$g(x)=ax^2-6x+15=-4x^2-6x+15$$

Then,

$$g'(x)=-8x-6$$

Setting $$g'(x)=0,$$

$$-8x-6=0$$

$$x=-\frac34$$

Also,

$$g''(x)=-8<0$$

Hence,

$$g(x)$$ has a local maximum at $$x=-\frac34$$