What is the answer to a and b please and thank you
1 answer:
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Answer:
The average value of over the interval
is
.
Step-by-step explanation:
Let suppose that function is continuous and integrable in the given intervals, by integral definition of average we have that:
(1)
(2)
By Fundamental Theorems of Calculus we expand both expressions:
(1b)
(2b)
We obtain the average value of over the interval
by algebraic handling:
The average value of over the interval
is
.
Answer:
The correct option is D. Discontinuity at (1, 7), zero at (negative four thirds, 0)
Step-by-step explanation:
To find the point of discontinuity :
Put the denominator equal to 0
⇒ x - 1 = 0
⇒ x = 1
Also, if the factor (x - 1) gets cancel, then it becomes a hole rather than a asymptote , ⇒ y = 3x + 4 at x = 1
⇒ y = 7
So, Point of discontinuity : (1, 7)
And the zero is : after cancelling the factor (x - 1) put the remaining factor = 0
⇒ 3x + 4 = 0
⇒ 3x = -4
⇒ x = negative four thirds ( zero of the function)
Therefore, The correct option is D. Discontinuity at (1, 7), zero at (negative four thirds, 0)
We want to determine the domain of 
any function of the form
is called an "exponential function",
the only condition is that b is positive and different from 1, and a is a nonzero real number.
The domain of such functions is all real numbers.
That is for any x, the expression <span>3(2^-x) "makes sense".
Answer: </span><span>The domain is all real numbers</span>
any function of the form
the only condition is that b is positive and different from 1, and a is a nonzero real number.
The domain of such functions is all real numbers.
That is for any x, the expression <span>3(2^-x) "makes sense".
Answer: </span><span>The domain is all real numbers</span>