The given function is f(x) = x⁶(x-1)⁵
The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
= x⁵(x-1)⁴(6x - 6 + 5x)
= x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.
Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
= (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
= (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
= 10x⁴(x-1)³(11x² - 12x + 3)
The sign of f''(x) at the critical values tell us the nature of the turning point.
f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.
The graphs shown below confirm these results.
Answer:
1. x = 2
2. x = 61/25
Step-by-step explanation:
Solve for x:
5 (x - 2) - 3 (2 - x) = 0
-3 (2 - x) = 3 x - 6:
3 x - 6 + 5 (x - 2) = 0
5 (x - 2) = 5 x - 10:
5 x - 10 + 3 x - 6 = 0
Grouping like terms, 5 x + 3 x - 10 - 6 = (3 x + 5 x) + (-6 - 10):
(3 x + 5 x) + (-6 - 10) = 0
3 x + 5 x = 8 x:
8 x + (-6 - 10) = 0
-6 - 10 = -16:
8 x + -16 = 0
Add 16 to both sides:
8 x + (16 - 16) = 16
16 - 16 = 0:
8 x = 16
Divide both sides of 8 x = 16 by 8:
(8 x)/8 = 16/8
8/8 = 1:
x = 16/8
The gcd of 16 and 8 is 8, so 16/8 = (8×2)/(8×1) = 8/8×2 = 2:
Answer: x = 2
_____________________________
Solve for x:
Solve for x:
3 (2 x - 7) + (7 x + 2)/3 = 0
Put each term in 3 (2 x - 7) + (7 x + 2)/3 over the common denominator 3: 3 (2 x - 7) + (7 x + 2)/3 = (9 (2 x - 7))/3 + (7 x + 2)/3:
(9 (2 x - 7))/3 + (7 x + 2)/3 = 0
(9 (2 x - 7))/3 + (7 x + 2)/3 = (9 (2 x - 7) + (7 x + 2))/3:
(9 (2 x - 7) + 2 + 7 x)/3 = 0
9 (2 x - 7) = 18 x - 63:
(18 x - 63 + 7 x + 2)/3 = 0
Grouping like terms, 18 x + 7 x - 63 + 2 = (18 x + 7 x) + (2 - 63):
((18 x + 7 x) + (2 - 63))/3 = 0
18 x + 7 x = 25 x:
(25 x + (2 - 63))/3 = 0
2 - 63 = -61:
(25 x + -61)/3 = 0
Multiply both sides of (25 x - 61)/3 = 0 by 3:
(3 (25 x - 61))/3 = 3×0
(3 (25 x - 61))/3 = 3/3×(25 x - 61) = 25 x - 61:
25 x - 61 = 3×0
0×3 = 0:
25 x - 61 = 0
Add 61 to both sides:
25 x + (61 - 61) = 61
61 - 61 = 0:
25 x = 61
Divide both sides of 25 x = 61 by 25:
(25 x)/25 = 61/25
25/25 = 1:
Answer: x = 61/25
Answer:
c = -54
Step-by-step explanation:
Solve for c:
(7 c)/8 - 3 (c/8 - 7) = -6
Put each term in c/8 - 7 over the common denominator 8: c/8 - 7 = c/8 - 56/8:
(7 c)/8 - 3 c/8 - 56/8 = -6
c/8 - 56/8 = (c - 56)/8:
(7 c)/8 - 3(c - 56)/8 = -6
(7 c)/8 - (3 (c - 56))/8 = (7 c - 3 (c - 56))/8:
(7 c - 3 (c - 56))/8 = -6
-3 (c - 56) = 168 - 3 c:
(7 c + 168 - 3 c)/8 = -6
7 c - 3 c = 4 c:
(4 c + 168)/8 = -6
Multiply both sides of (4 c + 168)/8 = -6 by 8:
(8 (4 c + 168))/8 = -6×8
(8 (4 c + 168))/8 = 8/8×(4 c + 168) = 4 c + 168:
4 c + 168 = -6×8
8 (-6) = -48:
4 c + 168 = -48
Subtract 168 from both sides:
4 c + (168 - 168) = -168 - 48
168 - 168 = 0:
4 c = -168 - 48
-168 - 48 = -216:
4 c = -216
Divide both sides of 4 c = -216 by 4:
(4 c)/4 = (-216)/4
4/4 = 1:
c = (-216)/4
The gcd of -216 and 4 is 4, so (-216)/4 = (4 (-54))/(4×1) = 4/4×-54 = -54:
Answer: c = -54
Answer: 2 nickels
Rachel has 20 coins composed of nickles, dimes and quarters
amounting to $3.70. Her quarters are twice the number of her dimes, the number
of nickels can be computed as follows:
If X= the number of
dimes at .10 each
Then 2x= the number
of quarters at .25 each and
20-(X-2X)= the number of nickels at .05 each
<span>
<span><span>
<span>
The equation will be:
X(.1)+2(X).25+(20-X-2X).10=3.70
</span>
</span>
<span>
</span>
<span>
<span>
.1X+2X(.25)+(20-X-2X).05=3.70
</span>
</span>
<span>
<span>
.1X+.5X+(20-3X).05=3.70
</span>
</span>
<span>
<span>
.6X+(1-.15X)=3.70
</span>
</span>
<span>
<span>
.6X+1-.15X=3.70
</span>
</span>
<span>
<span>
.45X=3.70-1
</span>
</span>
<span>
<span>
.45X=2.70
</span>
</span>
<span>
<span>
X=2.70/.45
</span>
</span>
<span>
<span>
X=6 number of dimes
</span>
</span>
<span>
<span>
2X=12 number of quarters
</span>
</span>
<span>
<span>
20-6-12=2 number of nickels
</span>
</span>
</span></span>