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Wewaii [24]
3 years ago
10

Determine whether the Mean Value Theorem applies to the function f(x)=2x^1/5 on the interval [-32, 32]

Mathematics
1 answer:
Anettt [7]3 years ago
5 0

Answer

given,

f(x) = 2 x^{1/5}

interval = [-32, 32]        

differentiating the given equation

f'(x) = \dfrac{d}{dx}(2 x^{1/5})

f'(x) = \dfrac{2}{5x^{4/5}}

hence, the above solution is not defined at x = 0

and x = 0 lie in the given interval i.e. [-32, 32]

so, at x = 0 the function is not differentiable.

Hence, mean value theorem does not apply to the given function.

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7. A, B, C, and D are positive integers. A,B,C forms an arithmetic sequence while B, C, D forms a
Stella [2.4K]

<u>Complete Question</u>

The positive integers A, B, and C form an arithmetic sequence while the integers B, C, and D form a geometric sequence. If (C/B) = (5/3), what is the smallest possible value of A + B + C + D?

Answer:

52

Step-by-step explanation:

If A, B, and C form an arithmetic progression

Their arithmetic mean, B=\dfrac{A+C}{2}

2B=A+C

C= 2B-A

B, C, D forms a  geometric sequence and Common ratio, r=C/B=5/3

The terms in the geometric sequence are:

B, B(\frac{5}{3} ), B(\frac{5}{3} )^2=B,  \frac{5B}{3} ,  \frac{25B}{9}

Therefore:

C=\frac{5B}{3}\\D= \frac{25B}{9}

So:

A, B, C, D=A, B,  \frac{5B}{3} ,  \frac{25B}{9}

From arithmetic sequence

Common difference,d=B - A = \frac{5B}{3} - B

2B -\frac{5B}{3}=A

(2 -\frac{5}{3})B=A\\(\frac{1}{3})B=A\\A=\frac{B}{3}\\

A, B,C, D =\frac{B}{3},\;B, \;\frac{5B}{3},\;\frac{25B}{9}

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A,B,C,D=3,9,15,25

So the smallest possible value for:

A+B+C+D = 3+9+15+25 = 52

3 0
3 years ago
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