What is the extreme f(x) 1/((x^2-10x+35)^3) ?

The extreme f(x) 1/((x^2-10x+35)^3) is Saddle(0,0)

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extreme f(x) 1/((x^2-10x+35)^3)

extreme points

f (x) \frac{1}{( x ^{2} - 10 x + 35 ) ^{3}}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial f ^{2}} = \frac{30 f ( x ^{3} - 8 x ^{2} - x + 70 )}{( x ^{2} - 10 x + 35 ) ^{5}}

D (f, x) = - \frac{25 ( - x ^{2} + 4 x + 7 ) ^{2}}{( x ^{2} - 10 x + 35 ) ^{8}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

f (x) = x^{2} - 10 x - 1, 2 \leq x \leq 6

f (x) = x^{3} + 15 x - 9

What is the extreme f(x) 1/((x^2-10x+35)^3) ?
The extreme f(x) 1/((x^2-10x+35)^3) is Saddle(0,0)