If-range-of-f-x-x-1-x-2-x-3-x-4-5-x-6-6-is-a-b-a-b-N-find-a-b-

Question Number 33036 by rahul 19 last updated on 09/Apr/18

I f r a n g e o f

f (x) = (x + 1) (x + 2) (x + 3) (x + 4) + 5

x \in [- 6, 6] i s [a, b], a, b \in N, f i n d a + b ?

Answered by MJS last updated on 09/Apr/18

we need the minimum and

maximum of f (x) in the given

interval

f (x) = x^{4} + 10 x^{3} + 35 x^{2} + 50 x + 29

f^{'} (x) = 4 x^{3} + 30 x^{2} + 70 x + 50 =

= 2 (2 x + 5) (x^{2} + 5 x + 5)

f^{'} (x) = 0

x_{1} = - \frac{5}{2} = - 2.5

x_{2} = - \frac{5}{2} - \frac{\sqrt{5}}{2} \approx - 3.618

x_{3} = - \frac{5}{2} + \frac{\sqrt{5}}{2} \approx - 1.382

f^{″} (x) = 12 x^{2} + 60 x + 70

f^{″} (x_{1}) < 0 \Rightarrow x_{1} is local maximum

f^{″} (x_{2}) > 0 \Rightarrow x_{2} is local minimum

f^{″} (x_{3}) > 0 \Rightarrow x_{3} is local minimum

f (x_{1}) = \frac{89}{16} = 5.5625 < f (- 6) < f (6)

f (x_{2}) = f (x_{3}) = 4

m a x (f (x)) = f (6) = 5045; x \in [- 6; 6]

m i n (f (x)) = f (- \frac{5}{2} \pm \frac{\sqrt{5}}{2}) = 4; x \in R

5045 + 4 = 5049

Commented by rahul 19 last updated on 09/Apr/18

i a m n o t a b l e t o u n d e r s t a n d l a s t 2 l i n e s;

w h a t i f 5.5625 > f (6) ? a n d i f i t l i e s

i n b e t w e e n f (- 6), f (6) ?

s i m i l a r l y w e h a v e n o t p u t f (- 6)

a s i t d o e s n o t g i v e m i n . v a l u e ?

w h y x \in [- 6, 6] a n d x \in R o t h e r t i m e ?

Commented by MJS last updated on 09/Apr/18

in this case, f (6) = 5045 is the

maximum for x \in [- 6; 6]

we found local maximum at

x = - \frac{5}{2} which c o u l d be the

maximum in [- 6; 6] but {it}^{'} s not

because f (- \frac{5}{2}) = \frac{89}{16} < 5045

so the maximum of f (x) im this

interval is 5045

the minimum is 4, {it}^{'} s even the

absolute minimum of f (x) for

x \in R \Rightarrow {it}^{'} s also the minimum

within [- 6; 6]

Commented by rahul 19 last updated on 09/Apr/18

t h a n k u s o m u c h s i r!

Commented by MJS last updated on 10/Apr/18

{you}^{'} re welcome

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