Compute the necessary values/derivatives of at
:
Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) at
by
###
Another way of doing this would be to solve for the coefficients in
by expanding the right hand side and matching up terms with the same power of .
Answer:
see explanation
Step-by-step explanation:
The nth term of a geometric sequence is
= a₁
where a₁ is the first term and r the common ratio
(1)
a₁ = 2 and r = =
= - 3 , then
= 2
(2)
a₁ = 5 and r = = 2 , then
= 5
The easiest answer I know is if you swap the s with
s, then there will be two equations called parabolas. They open upward if the value is positive and open downward if the value is negative. If the exponent is 2, then it's a parabola. If it's 3, then it's a cubic function, because cube is for "3". Square is for "2". If you look at both parabolas, you can see that
is wider than
.
So the best answer I can think of is She made a mistake because the parabola on the 10x² graph is wider than the (5x)² graph. Also, it's because 5x and 10x are not the same thing. When an x value increases, the graph is narrower.