Question

F(x) = 3x + x3

x > 0
a) Differentiate to find f'(x)
Given that f'(x) = 15,
b) Find the value of x​

Answer 1

Please check the explanation.

Explanation:

Given

f(x) = 3x + x³

Taking differentiate

`(d)/(dx)\left(3x+x^3\right)`

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

`=(d)/(dx)\left(3x\right)+(d)/(dx)\left(x^3\right)`

solving

`(d)/(dx)\left(3x\right)`

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

`=3(d)/(dx)\left(x\right)`

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

`=3\cdot \:1`

`=3`

now solving

`(d)/(dx)\left(x^3\right)`

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

`=3x^(3-1)`

`=3x^2`

Thus, the expression becomes

`(d)/(dx)\left(3x+x^3\right)=(d)/(dx)\left(3x\right)+(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=√(f\left(a\right)),\:\:-√(f\left(a\right))`

`x=√(4),\:x=-√(4)`

`x=2,\:x=-2`

Thus, the value of x​ will be:

`x=2,\:x=-2`