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sergeinik [125]
3 years ago
9

PLEASE HELP ME !!

Mathematics
1 answer:
julsineya [31]3 years ago
6 0
Second or third one not sure
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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
4 years ago
Simplify:<br><br> Cube root of (128a^13b^6)
Pavlova-9 [17]
\sqrt[3]{128a^{13}b^{6}} \\\sqrt[3]{64a^{12}b^{6} * 2a} \\\sqrt[3]{64a^{12}b^{6}}\sqrt[3]{2a} \\4a^{4}b^{2}\sqrt[3]{2a}
7 0
3 years ago
Ill mark brainlist plss help
Gelneren [198K]

Answer:

quadrilateral, parallelogram, rhombus

8 0
3 years ago
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You are designing a container in the shape of a cylinder. The radius is 6 inches. You want the container to hold at least 324π32
Volgvan
The least possible height is the height of the cylinder when the volume is 324π cubic inches. To determine that height, we use the formula for the volume of a cylinder. We do as follows:

Volume = πr^2h
324π = π(6)^2(h)
h = 9 inches
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UV has endpoints at U(8, 2) and V(8,0). Find the midpoint M of UV.
Marat540 [252]
The midpoint here which is = to M is (8, 1). Hope this helps
5 0
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