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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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-12 (-5/6) (-2/15) How do I solve this question I am stuck on an assignment
koban [17]

If a number in parenthesis is directly next to a number without, then the two numbers need multiplied.

-12(5/6)(-2/15) is an example of a double multiplication. But we'll multiply -12 and 5/6 first, then multiply the product by -2/15.

-12x5/6=14 2/5

14 2/5x-2/15=-1 23/25

---

hope it helps

sorry if it doesn't

6 0
3 years ago
Work out the value of x.
Bas_tet [7]

Answer:

x = 33

Step-by-step explanation:

Sum of the interior angles of a 5-sided polygon = 540°

Therefore all the interior angles of the given polygon would all be equal to 540°.

Thus:

(4x - 25)° + (3x + 25)° + (2x - 3)° + (4x)° + (3x + 15)° = 540°

Solve for x

4x - 25 + 3x + 25 + 2x - 3 + 4x + 3x + 15 = 540

Combine like terms

4x + 3x + 2x + 4x + 3x - 25 + 25 - 3 + 15 = 540

16x + 12 = 540

16x = 540 - 12

16x = 528

Divide both sides by 16

x = 33

5 0
3 years ago
What is -h+4 = -h+9????
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7 0
3 years ago
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