Answer:
Please check the explanation.
Step-by-step explanation:
Given
f(x) = 3x + x³
Taking differentiate
[tex]\frac{d}{dx}\left(3x+x^3\right)[/tex]
[tex]\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'[/tex]
[tex]=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)[/tex]
solving
[tex]\frac{d}{dx}\left(3x\right)[/tex]
[tex]\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'[/tex]
[tex]=3\frac{d}{dx}\left(x\right)[/tex]
[tex]\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1[/tex]
[tex]=3\cdot \:1[/tex]
[tex]=3[/tex]
now solving
[tex]\frac{d}{dx}\left(x^3\right)[/tex]
[tex]\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}[/tex]
[tex]=3x^{3-1}[/tex]
[tex]=3x^2[/tex]
Thus, the expression becomes
[tex]\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)[/tex]
[tex]=3+3x^2[/tex]
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
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{4},\:x=-\sqrt{4}[/tex]
[tex]x=2,\:x=-2[/tex]
Thus, the value of x will be:
[tex]x=2,\:x=-2[/tex]
