y(r) = 4 r^2 + 9 = u
y(c) = 4 c^2 + 9 = g
----------------------subtract
y(r)-y(c) = 4(r^2-c^2) = u-g =delta y
delta x = (r-c)
delta y/delta x
= 4(r^2-c^2)/(r-c)
= 4(r+c)
I do not understand B
how can you solve average rate of change in f(x)=4x^2+9 in following points?
A) (c,g) (r,u)
B) 2x+h
6 answers
B is (x+h) and f(x+h) which I believe equals 2x+h
Oh, I do not know what your belief means but I know what
x+h and f(x+h) means.
They are trying to lead you to calculus
x+h and f(x+h) means.
They are trying to lead you to calculus
f(x+h) = 4(x+h)^2 + 9
f(x) = 4 x^2 + 9
-----------------subtract
delta y =4[x^2 + 2 x h + h^2 -x^2]
= 4 (2 x h + h^2)
delta x = h
so
delta y/delta x = 4(2x + h)
END OF YOUR PROBLEM
Oh my look what we have here as h ---> 0 !!
delta y/delta x --> 4 (2x)
That is called the DERIVATIVE !!!
that is
f(x) = 4 x^2 + 9
-----------------subtract
delta y =4[x^2 + 2 x h + h^2 -x^2]
= 4 (2 x h + h^2)
delta x = h
so
delta y/delta x = 4(2x + h)
END OF YOUR PROBLEM
Oh my look what we have here as h ---> 0 !!
delta y/delta x --> 4 (2x)
That is called the DERIVATIVE !!!
that is
That is important.
No I do not. I haven't taken calculus at all. I'm barely in college algebra.