Let 'a' be a real number such that the function f(x) = ax2 + 6x - 15, x [] R is increasing…

Question

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 :

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 $$x = {{ - B} \over {2A}} =  - {{( - 6)} \over {2( - 4)}}$$

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

Solutions · 1