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:

$f'(c) = \frac{f(b)-f(a)}{b-a}$

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

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