If the interval (a, ∞) describes all values of x for which the graph of 4/x^2-6x+9 is decreasing, what is the value of a?
2 answers:
Answer:
a = - 1.1
Step-by-step explanation:
Condition for a single variable function f(x) to be decreasing is f'(x) < 0.
So, in our case, .
Then, differentiating with respect to x on both sides we get,
⇒
⇒
{Dividing both sides with - 1 and hence the inequality sign changes}
⇒
⇒ 8 > - 6x³ {Multiplying both sides with x³}
⇒ 6x³ > - 8 {Interchanging the sides}
⇒
⇒ x > - 1.1
Therefore, a = - 1.1. (Answer)
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Answer:
The Taylor series of f(x) around the point a, can be written as:
Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as:
Answer:
a.) f(x) = -⅙(x+3)²+6
Step-by-step explanation:
The maximum value, our vertex, is at point (-3,6).
We can insert this value into the vertex form of a quadratic function and then solve for a as follows...
a equals -1/6... We can input this into the original equation we used...
f(x) = -1/6(x+3)^2+6
Good luck on the bellwork ;)