Answer:
Here we just want to find the Taylor series for f(x) = ln(x), centered at the value of a (which we do not know).
Remember that the general Taylor expansion is:
for our function we have:
f'(x) = 1/x
f''(x) = -1/x^2
f'''(x) = (1/2)*(1/x^3)
this is enough, now just let's write the series:
This is the Taylor series to 3rd degree, you just need to change the value of a for the required value.
what is f(-3) is another way to say "what is the value of f(x) when x = -3?", well for that we can simply look at the piece-wise function graphed, Check the picture below.
im not sure this is right
g'(x) = 6b(-5x + 1)^5 (-5)
g'(x) = -30b(-5x +1)^5
g''(x) = -30b(5)(-5x + 1)^4 (-5)
g''(x) = 750b (-5x +1)^4
g(x) = b(−5x + 1)6 − a
when
g(-x) = b(5x +1)6 - a
g'(x) = -30b(-5x +1)^5 = 0
-5x +1 = 0
x = 15
Answer:
2) y=5x+13
When in doubt, replace the variable with a number to see if it makes sense
Answer:
p<-3/8
Step-by-step explanation:
