now, what is the value of the derivative at any extrema on the original function? well, by definition, if the original function has an extrema anywhere, the derivative will be the derivative of a horizontal tangent line to it, and by definition it'll be 0.
Answer:
Circle
Step-by-step explanation:
Answer:
-2
The problem:
y=f(x)=x^2+x
find the slope of the secant line joining (-3,f(-3)) and (0,f(0))
Step-by-step explanation:
We need to first find the values that correspond to f(-3) and f(0).
f(-3) can be found by replacing the x's in f(x)=x^2+x with (-3):
f(-3)=(-3)^2+(-3)
f(-3)=9+-3
f(-3)=6
f(0) can be found by replacing the x's in f(x)=x^2+x with (0):
f(0)=(0)^2+(0)
f(0)=0+0
f(0)=0
So we want to find the slope of the line going through the points:
(-3,6) and (0,0)
Line them up-doesn't matter which point goes on top:
(-3,6)
(0,0)
---------Now subtract!
-3 , 6
The rise goes over the run so the slope is 6/-3 which simplifies to -2.
