Question

If the interval (a, infinity) describes all values of x for which the graph of f(x)=4/x^2-6x+9 is decreasing, what is the value of a

Answer 1

Answer:

The answer is "3".

Step-by-step explanation:

$\texttt{Given graph equation: } \\\\\bold{\Rightarrow f(x)=\frac{4}{x^2-6x+9} }$

$\bold{\ interval: (a,\infty)}$

differentiate the function f(x):

$\Rightarrow f'(x) = \frac{d }{dx}(\frac{4}{x^2-6x+9})$

Formula:

$\bold{\frac{d}{dx} \frac{v}{u}= \frac{u \frac{d}{dx} v- v\frac{d}{dx}u }{u^2}}$

$\Rightarrow f'(x) = \frac{d }{dx}(\frac{4}{x^2-6x+9})\\\\\\\Rightarrow f'(x) = \frac{d }{dx}(\frac{(x^2-6x+9) \frac{d}{dx} 4- 4\frac{d}{dx}(x^2-6x+9) }{(x^2-6x+9)^2})\\\\\\\Rightarrow f'(x) = \frac{d }{dx}(\frac{(x^2-6x+9) \times 0 - 4(2x-6) }{(x^2-6x+9)^2})\\\\\\\Rightarrow f'(x) = \frac{- 4(2x-6)}{(x^2-6x+9)^2}\\\\\\\Rightarrow \frac{- 4(2x-6)}{(x^2-6x+9)^2}=0\\\\\Rightarrow - 4(2x-6)=0\\\\\Rightarrow 8x-24=0\\\\\Rightarrow 8x=24\\\\\Rightarrow x=\frac{24}{8}\\\\\Rightarrow x=3$

since the value of x in: $(-3 ,\infty) \ \ and \ \ (3, \infty)$ and $f'(x) <= 0$. So, the value of a is 3

Answer 2

Answer:

3

Step-by-step explanation: