Consider the function f(x)=2x^3−15x^2−36x+8 on the interval [−6,8].

The average or mean slope of the function on this interval is equal to 38

By the Mean Value Theorem, we know there exists a c in the open interval (−6,8) such that f′(c) is equal to this mean slope. For this problem, there are two values of c that work:

The smaller one c1=

The larger one c2=

Question

Consider the function f(x)=2x^3−15x^2−36x+8 on the interval [−6,8].

The average or mean slope of the function on this interval is equal to  38

By the Mean Value Theorem, we know there exists a c in the open interval (−6,8) such that f′(c) is equal to this mean slope. For this problem, there are two values of c that work:

The smaller one c1=

The larger one c2=

myininaya · Answer

$$\frac{f(-6)-f(8)}{-6-8}=6c^2-30c-36$$
solve for c

anonymous · Answer

I got -1 and 6 but my online homework said this is wrong.... please help

myininaya · Answer

$$\frac{-748-(-216)}{-14}=6c^2-30c-36$$

$$38=6c^2-30c-36 $$

$$6c^2-30c-74=0$$

$$3c^2-15c-37=0$$

anonymous · Answer

I got -1.53112887414927 and 6.53112887414927	but those answers were wrong too... :/

myininaya · Answer

$$c=\frac{15 \pm \sqrt{(15)^2-4(3)(-37)}}{2(3)}=\frac{15 \pm \sqrt{669}}{6}$$

myininaya · Answer

$$c_1=\frac{15- \sqrt{669}}{6}, c_2=\frac{15+\sqrt{669}}{6}$$

anonymous · Answer

omg! thank you! ^-^