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Oksana_A [137]
3 years ago
14

(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 thi

s 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:
Mathematics
1 answer:
OlgaM077 [116]3 years ago
7 0

Answer: c = 4.97 and c = -1.97

Step-by-step explanation: <u>Mean Value Theorem</u> 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) = \frac{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 = \frac{100-113}{9-(-4)}

6c^{2}-18c-60 = -1

6c^{2}-18c-59 = 0

Resolving through Bhaskara:

c = \frac{18+\sqrt{1740} }{12}

c = \frac{18+41.71 }{12} = 4.97

c = \frac{18-41.71 }{12} = -1.97

<u>Both values of c exist inside the interval [-4,9], so both values are mean slope: c = 4.97 and c = -1.97</u>

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