Answer:
3 okie
Step-by-step explanation:
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:
y = 5^x
Step-by-step explanation:
y= b*(a)^x + c
c could = 1 but then you would not have an exponential function. c = 0 because the graph follows the x axis up until x = -2. Suppose c = 1. The the graph would follow y = 1 up until x = - 2
When x = 0, y = 1 which means that b. If b is anything but 0 or 1 then the y intercept would be stretched to a different place. If be = 0 then y would = 0.
So the graph is of the form y = a^x
Now when x = 0 the graph, the y intercept is y = a^0 or y = 1 So the y intercept is (0,1)
Now the next point is thing to solve for is a.
When x = 1, y = 5 (read the graph)
y = a^x
5 = a^1
5 = a because a^1 is a.
Answer
y = 5^x.
5 = a^1
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