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topjm [15]
3 years ago
8

Mum is 25 years older than her son Max, but she is 1 year younger than

Mathematics
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

17

Step-by-step explanation:

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3 years ago
Complete parts a through c for the given function.
victus00 [196]

Answer:

a. The critcal points are at

x=0,-5,3

b. Then, x = -5   is a maximum and x=3 is a minimum

c. The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

Step-by-step explanation:

(a)

Remember that you need to find the points where

f'(x)=0

Therefore you have to solve this equation.

20x^4  + 40x^3 - 300x^2 = 0

From that equation you can factor out    20x^2  and you would get

20x^2 (  x^2  +2x - 15)  = 0

And from that you would have   20x^2 = 0  , so x = 0.

And you would also have  x^2 +2x-15 = 0.

You can factor that equation as    x^2 +2x -15 = (x+5)(x-3) = 0

Therefore   x=-5 ,   x=3.

So the critcal points are at

x=0,-5,3

b.  

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\

Then, x = -5   is a maximum and x=3 is a minimum

c.

The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

6 0
3 years ago
Read 2 more answers
Help anyone ??? Please
lana [24]
Angle F is equal to angle G so 5x + 18 = 7x-12 so you would solve for x which would equal 15
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3 years ago
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