What is the extreme f(x)=-(15)/(x^2-19x+16) ?

The extreme f(x)=-(15)/(x^2-19x+16) is Minimum(19/2 , 20/99)

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extreme f(x)=-(15)/(x^2-19x+16)

extreme

Minimum (\frac{19}{2}, \frac{20}{99})

Solution steps

f^{'} (x) = \frac{15 ( 2 x - 19 )}{( x ^{2} - 19 x + 16 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < \frac{19 - 3 33}{2},

Decreasing

: \frac{19 - 3 33}{2} < x < \frac{19}{2},

Increasing

: \frac{19}{2} < x < \frac{19 + 3 33}{2},

Increasing

: \frac{19 + 3 33}{2} < x < \infty

Plug

x = \frac{19}{2}

into

- \frac{15}{x ^{2} - 19 x + 16} : \frac{20}{99}

Minimum (\frac{19}{2}, \frac{20}{99})

What is the extreme f(x)=-(15)/(x^2-19x+16) ?
The extreme f(x)=-(15)/(x^2-19x+16) is Minimum(19/2 , 20/99)