Consider the function f(x)=2x^3-6x^2-90x+10 on the interval [-5,7]

Question

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

By the Mean Value Theorem, we know there exists a c in the open interval (-5,7) 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= ____

Answer 1

f'(c) = 6c² -12c -90 = f(7) -f(-5)) /(7 - (-5)) f(7)= -228 and f(-5) = 60 so f'(c)=(-228 -60 )/12 = -24 so now

6c² -12c -90 = -24 and 6c² -12c -66 =0 or c² -2c -11 =0 complete the square to get (c -1)² = 12 and

c-1 = ± 2√3 so the smaller c = 1 - 2√3 and the larger c = 1 + 2√3 both of these are in the interval (-5,7)

Answer 2

The Mean Value Theorem (MVT) has requirements:

1. The function must be continuous on the closed interval.
2. The function must be differentiable on the open interval.

Since the function is a polynomial, both requirements are fulfilled.

What does MVT state? Once the function fulfills the requirements on the given interval, then:

f ' (c) = [ f (b) − f (a) ] / [b − a]

Essentially, there must be at least one point on the interval such that the derivative at that point is equivalent the slope of the secant line connecting the endpoints of the interval.

f (x) = 2x³ − 6x² − 90x + 10

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Turning our attention to the LEFT−HAND SIDE of the equation, f ' (c). Take the derivative of f.

f (x) = 2x³ − 6x² − 90x + 10

Using the Power Rule:

f ' (x) = 6x² − 12x − 90

Substituting in c for x:

f ' (c) = 6c² − 12c − 90

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Turning to the RIGHT−HAND SIDE of the MVT:

[ f (b) − f (a) ] / [b − a]

Calculate f (b) and f (a), where a = −5 and b = 7:

f (−5) = 2(−5)³ − 6(−5)² − 90(−5) + 10 = 60

f (7) = 2(7)³ − 6(7)² − 90(7) + 10 = −228

Substitute the values into the right-hand side of the equation to get the slope of the secant line:

[ f (b) − f (a) ] / [b − a]

= (-228 − 60) / (7 − (-5))

= −288 / 12

= -24

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Plug in both results into the MVT. Substituting the values and simplifying:

f ' (c) = [ f (b) − f (a) ] / [b − a]

6c² − 12c − 90 = −24

6c² − 12c − 66 = 0

6 (c² − 2c − 11) = 0

c² − 2c − 11 = 0

Using the Quadratic Formula, we get two values.

c₁ = 1 − 2√3 ≈ −2.464 (low value)

c₂ = 1 + 2√3 ≈ 4.464 (high value)

Once the potential 'c' points are calculated, we have to ensure that the final solution set only includes values within the closed interval. Sometimes, solutions generated by MVT may not be included in the domain of the original problem statement.

In this case, both c₁ and c₂ can be found within the closed interval given in the problem statement [−5, 7]..