Answer: c = 4.97 and c = -1.97
Step-by-step explanation: Mean Value Theorem states if a function f(x) is continuous on interval [a,b] and differentiable on (a,b), there is at least one value c in the interval (a<c<b) such that:
[tex]f'(c) = \frac{f(b)-f(a)}{b-a}[/tex]
So, for the function f(x) = [tex]2x^{3}-9x^{2}-60x+1[/tex] on interval [-4,9]
[tex]f'(x) = 6x^{2}-18x-60[/tex]
f(-4) = [tex]2.(-4)^{3}-9.(-4)^{2}-60.(-4)+1[/tex]
f(-4) = 113
f(9) = [tex]2.(9)^{3}-9.(9)^{2}-60.(9)+1[/tex]
f(9) = 100
Calculating average:
[tex]6c^{2}-18c-60 = \frac{100-113}{9-(-4)}[/tex]
[tex]6c^{2}-18c-60 = -1[/tex]
[tex]6c^{2}-18c-59 = 0[/tex]
Resolving through Bhaskara:
c = [tex]\frac{18+\sqrt{1740} }{12}[/tex]
c = [tex]\frac{18+41.71 }{12}[/tex] = 4.97
c = [tex]\frac{18-41.71 }{12}[/tex] = -1.97
Both values of c exist inside the interval [-4,9], so both values are mean slope: c = 4.97 and c = -1.97
