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klasskru [66]
3 years ago
15

Every number between -1 and 1 including -1 and 1

Mathematics
1 answer:
Shkiper50 [21]3 years ago
3 0
Isnt the first one zero becuse the number between -1 and 1 coupd only be zero
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If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
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\bf f(x)=y=2x+sin(x)
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inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
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\textit{now, the "y" in the inverse, is really just g(x)}
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\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
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now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
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g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
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Answer:

44degree................

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garik1379 [7]

Answer:

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Step-by-step explanation:

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