Answer:
We say that f(x) has an absolute (or global) minimum at x=c if f(x)≥f(c) f ( x ) ≥ f ( c ) for every x in the domain we are working on. We say that f(x) has a relative (or local) minimum at x=c iff(x)≥f(c) f ( x ) ≥ f ( c ) for every x in some open interval around x=c .
Answer:
see explanation
Step-by-step explanation:
The common difference d of an arithmetic sequence is
d =
-
=
- 
Substitute in values and solve for k, that is
5k - 1 - 2k = 6k + 2 - (5k - 1)
3k - 1 = 6k + 2 - 5k + 1
3k - 1 = k + 3 ( subtract k from both sides )
2k - 1 = 3 ( add 1 to both sides )
2k = 4 ⇒ k = 2
--------------------------------------------------------
The n th term of an arithmetic sequence is
=
+ (n - 1)d
= 2k = 2 × 2 = 4 and
d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5
Hence
= 4 + (7 × 5) = 4 + 35 = 39
It is 6 because 1000 would not be the appropiate answer
Answer:
g(x) = sinh^-1 ( ln(7x^6 +3) / sqrt( 8+cot( x^( 3+x))))
Step-by-step explanation:
Using the fundamental theorem of calculus
Taking the derivative of the integral gives back the function
Since the lower limit is a constant when we take the derivative it is zero
d/dx 
g(t) = sinh^-1 ( ln(7t^6 +3) / sqrt( 8+cot( t^( 3+t))))
Replacing t with x
g(x) = sinh^-1 ( ln(7x^6 +3) / sqrt( 8+cot( x^( 3+x))))