Answer:
Step-by-step explanation:
According to Rolle's Theorem, if f(a) = f(b) in an interval [a, b], then there must exist at least one <em>c</em> within (a, b) such that f'(c) = 0.
We are given that g(5) = g(8) = -9. Then according to Rolle's Theorem, there must be a <em>c</em> in (5, 8) such that g'(c) = 0.
So, differentiate the function. We can take the derivative of both sides with respect to <em>x: </em>
<em />![\displaystyle g'(x) = \frac{d}{dx}\left[ -7x^3 +91x^2 -280x - 9\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20g%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%20-7x%5E3%20%2B91x%5E2%20-280x%20-%209%5Cright%5D)
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Differentiate:
Let g'(x) = 0:
Solve for <em>x</em>. First, divide everything by negative seven:
Factor:
<h3>
</h3>
Zero Product Property:
Solve for each case. Hence:
Since the first solution is not within our interval, we can ignore it.
Therefore:
Step-by-step explanation:
When x = 0,
sin3x / 6x = 0 / 0, which is an indeterminate form.
Hence we use L'Hospital's rule:
d/dx (sin3x) = 3cos3x
d/dx (6x) = 6
Now we have 3cos3x / 6 or 0.5cos3x.
When x = 0, 0.5cos3x = 0.5(1) = 0.5.
Hence the limit is 0.5.
Answer:
P ( 1.2 < X < 2.1 ) = 0.3
Step-by-step explanation:
Given:
Uniform distribution over interval (0,3) can be modeled by a probability density function f(x)
f(x) = 1 / (b - a)
Where a < x < b is the domain at which function is defined:
f(x) = 1 / (3) = 1 / 3
Where, X - U ( u , δ )
u = ( a + b ) / 2 = (0 +3) / 2 = 1.5
δ = ( b - a ) / sqrt (12) = (3 - 0) / sqrt (12) = 0.866
Hence,
X - U ( 1.5 , 0.866 )
There-fore calculating P ( 1.2 < X < 2.1 ):
Where, a = 1.2 and b = 2.1
P ( 1.2 < X < 2.1 ) = x / 3 |
P ( 1.2 < X < 2.1 ) = 2.1 /3 - 1.2 / 3 = 0.3
Answer: P ( 1.2 < X < 2.1 ) = 0.3
Answer:
5
Step-by-step explanation:
