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Taya2010 [7]
3 years ago
13

Can some one help mẹ pls....

Mathematics
1 answer:
statuscvo [17]3 years ago
4 0

Answer:

h(2) or h(g(4)) = -28

Step-by-step explanation:

g(t) = t - 2,

g(4) = 4 - 2,

g(4) = 2

h(t) = -3t³ - 2t,

h(2) = -3(2)³ - 2(2),

h(2) = -24 - 4,

h(2) or h(g(4)) = -28

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7 0
3 years ago
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}

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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)

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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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