Question

(1 point) Consider the function f(x)=2x3−9x2−60x+1 on the interval [−4,9]. Find the average or mean slope of the function on this interval. Average slope: By the Mean Value Theorem, we know there exists at least one value c in the open interval (−4,9) such that f′(c) is equal to this mean slope. List all values c that work. If there are none, enter none . Values of c:

Answer 1

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:

`f'(c) = (f(b)-f(a))/(b-a)`

So, for the function f(x) =
`2x^(3)-9x^(2)-60x+1` on interval [-4,9]

`f'(x) = 6x^(2)-18x-60`

f(-4) =
`2.(-4)^(3)-9.(-4)^(2)-60.(-4)+1`

f(-4) = 113

f(9) =
`2.(9)^(3)-9.(9)^(2)-60.(9)+1`

f(9) = 100

Calculating average:

`6c^(2)-18c-60 = (100-113)/(9-(-4))`

`6c^(2)-18c-60 = -1`

`6c^(2)-18c-59 = 0`

Resolving through Bhaskara:

c =
`(18+√(1740) )/(12)`

c =
`(18+41.71 )/(12)` = 4.97

c =
`(18-41.71 )/(12)` = -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