So it tells us that g(3) = -5 and g'(x) = x^2 + 7.
So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))
now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16
Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84
So we have our linear approximation for the two.
If you wanted to, you could check your answer by finding g(x). Since you know g'(x), take the antiderivative and we will get
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35
So just to check our linear approximations use that to find g(2.99) and g(3.01)
g(2.99) = -5.1597
g(3.01) = -4.8397
So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer. Not a bad method if you ever need to use it.
The square of a prime number is not prime.
a) let x ∈ R, If x ∈ {prime numbers}, then 
∉{prime numbers}
there says that if x is a real and x is in the set of the prime numbers, then the square of x isn't in the set of prime numbers.
b) Prove or disprove the statement.
ok, if x is a prime number, then x only can be divided by himself. Now is easy to see that = x*x can be divided by himself and x, then x*x is not a prime number, because can be divided by another number different than himself
Answer:
y = -2x - 8
Step-by-step explanation:
The formula for a slope line is
y = mx + b
"m" is the slope that's given in the problem, and b is simply the y value of the y-intercept, which is also given.
I hope this helps!
An infinite amount seeing as you can have infinite of each number
We will use u-substitute:
Then for substitution:dx=u du. and integral becomes:
=u-ln(u+1)=
. Now we will change the values of limits:
=4-ln(5)-2+ln(3)=
2+ln(0,6)=2-0.51=1.49
