To find a number 10 times smaller than 62 simply move the decimal point one place to the left.
The decimal point in 62 is at the end of the number. 62. becomes 6.2 if we move the decimal one place back.
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(a) When f is increasing the derivative of f is positive.
f'(x) = 15x^4 - 15x^2 > 0
15x^2(x^2 - 1)> 0
x^2 - 1 > 0 (The inequality doesn't flip sign since x^2 is positive)
x^2 > 1
Then f is increasing when x < -1 and x > 1.
(b) The f is concave upward when f''(x) > 0.
f''(x) = 60x^3 - 30x > 0
30x(2x^2 - 1) > 0
x(2x^2 - 1) > 0
x(x^2 - 1/2) > 0
x(x - 1/sqrt(2))(x + 1/sqrt(2)) > 0
There are four regions here. We will check if f''(x) > 0.
x < -1/sqrt(2): f''(-1) = -30 < 0
-1/sqrt(2) < x < 0: f''(-0.5) = 7.5 > 0
0 < x < 1/sqrt(2): f''(0.5) = -7.5 < 0
x > 1/sqrt(2): f''(1) = 30 > 0
Thus, f''(x) > 0 at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
Therefore, f is concave upward at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
(c) The horizontal tangents of f are at the points where f'(x) = 0
15x^2(x^2 - 1) = 0
x^2 = 1
x = -1 or x = 1
f(-1) = 3(-1)^5 - 5(-1)^3 + 2 = 4
f(1) = 3(1)^5 - 5(1)^3 + 2 = 0
Therefore, the tangent lines are y = 4 and y = 0.
Answer:
2441.95 years
Step-by-step explanation:
We can model this exponencial function as:
P = Po * (1+r)^(t/n)
Where P is the final value, Po is the inicial value, r is the rate, t is the time and n is the period of half-life.
In this case, we have that P/Po = 100% - 25.5% = 74.5% = 0.745, r = -0.5 and n = 5750, so we have that:
0.745 = (1 - 0.5)^(t/5750)
Step 1: log in both sides:
log(0.745) = (t/5750) * log(0.5)
Step 2: isolate t
t = 5750*log(0.745)/log(0.5) = 2441.95 years
