Question

Consider the function f(x) = 3 -2 x^2 on the interval [ -5 , 3 ].
(A) Find the average or mean slope of the function on this interval, i.e.
\displaystyle{\frac { f(3) - f(-5) }{ 3 - (-5) } = }
4

(B) By the Mean Value Theorem, we know there exists a c in the open interval (-5, 3) such that f'( c) is equal to this mean slope. For this problem, there is only one c that works. Find it.
c=__????

Answer 1

at 3
y = 3 - 2(9) = -15
at -5
y = 3 - 2(25) = -47
-15 + 47 = 32
32/(3+5) = 4 agree

dy/dx = -4 x
at our spot
-4x = 4
so x = -1 where the slope is 4