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3 years ago

Hi guys can someone pls help me with this ASAP

Mathematics

1 answer:

Gennadij [26K]3 years ago

8 0

Answer:

the second one

Step-by-step explanation:

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3 years ago

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F(x) = 3x + x3

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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

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

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

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$=3x^2$

Thus, the expression becomes

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

$=3+3x^2$

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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3 years ago

Find the zeros of the function. f(x) = 6x3 + 18x2 + 12x

marin [14]

Hello,

f(x)=6x^3+18x²+12x=6x(x²+3x+2)
=6x(x²+x+2x+2)=6x((x(x+1)+2(x+1))
=6x(x+1)(x+2)

Zeros are 0,-1,-2

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3 years ago

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