What is the extreme f(x)=-(13)/(x^2-5x+35) ?

The extreme f(x)=-(13)/(x^2-5x+35) is Minimum(5/2 ,-52/115)

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extreme f(x)=-(13)/(x^2-5x+35)

extreme points

f (x) = - \frac{13}{x ^{2} - 5 x + 35}

Minimum (\frac{5}{2}, - \frac{52}{115})

Solution steps

f^{'} (x) = \frac{13 ( 2 x - 5 )}{( x ^{2} - 5 x + 35 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < \frac{5}{2},

Increasing

: \frac{5}{2} < x < \infty

Plug

x = \frac{5}{2}

into

- \frac{13}{x ^{2} - 5 x + 35} : - \frac{52}{115}

Minimum (\frac{5}{2}, - \frac{52}{115})

f (x) = \frac{3 x ^{2} - 4 x y + 2 y ^{2}}{4 x + 9 y ^{2}}

(x - 2)^{3} + 3

f (x) = ∣ x + 4 ∣, - 1 \leq x \leq 6

f (x) = - 2 x^{3} + 27 x^{2} - 84 x, 1 \leq x \leq 8

f (x) = 2 x^{3} - 4

What is the extreme f(x)=-(13)/(x^2-5x+35) ?
The extreme f(x)=-(13)/(x^2-5x+35) is Minimum(5/2 ,-52/115)