Raymond B. answered 04/07/23
Math, microeconomics or criminal justice
f(x)=7-48x+4x^2
f'(x) = -48 +8x
f(5) = 7-48(5)+4(5)^2= 107-240=-133
f(7)= 7-48(7)+4(7^2)= 7+196-336= -133
secant line has slope=0
f'(c)=-48+8c=0
8c=48
c=48/8
c =6
f(x)= 7-48x+4x^2= 4(x^2-12x+36)+7-4(36)
=4(x-6)^2 -137
f(5)= -133
f(7)=-133
upward opening parabola with vertex (6, -137)
which is continuous and differentiable on the interval 5 to 7