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Keith_Richards [23]
3 years ago
7

What is 1-1 please help

Mathematics
2 answers:
Tju [1.3M]3 years ago
7 0
Your answer going to be 0
vladimir1956 [14]3 years ago
4 0

Answer:

0

Step-by-step explanation:

1-1=0

It's common knowledge that 1-1=0 so I dont know how to explain it any further

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Step-by-step explanation:

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F(x) = 3x + x3
____ [38]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

\frac{d}{dx}\left(3x+x^3\right)

\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'

=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

solving

\frac{d}{dx}\left(3x\right)

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'

=3\frac{d}{dx}\left(x\right)

\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1

=3\cdot \:1

=3

now solving

\frac{d}{dx}\left(x^3\right)

\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}

=3x^{3-1}

=3x^2

Thus, the expression becomes

\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

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Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value  f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 =  3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

Thus, the value of x​ will be:

x=2,\:x=-2

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