Answer:
C. [tex]x < -3[/tex] or [tex]x > 1[/tex].
Step-by-step explanation:
Let [tex]f(x) = x^{3}+3\cdot x^{2}-9\cdot x +7[/tex], the first derivate of the function is:
[tex]f'(x) = 3\cdot x^{2}+6\cdot x -9[/tex] (1)
By the Quadratic Formula, we determine the roots of this polynomial:
[tex]x_{1} = 1[/tex], [tex]x_{2} = -3[/tex]
And the factorized form of the polynomial is:
[tex](x-1)\cdot (x+3) > 0[/tex]
By law of Signs, the polynomial is increasing when [tex]x-1 > 0[/tex] and [tex]x +3 > 0[/tex] or [tex]x-1 < 0[/tex] and [tex]x+3 < 0[/tex], which means that:
[tex]x < -3[/tex] or [tex]x > 1[/tex]
Therefore, the right answer is C.
Answer:
The correct option is an option (c).
Given:
The given function is,
[tex]f(x)=x^{3}+3x^{2}-9x+7[/tex]
Computation:
The function is increasing when [tex]f'(x)>0[/tex].
Now, differentiating the given function with respect to [tex]x[/tex] up to first order we get,
[tex]f'(x)=3x^{2}+3(2x)-9(1)+0\\f'(x)=3x^{2}+6x-9\\=3(x^{2}+2x-3)\\=3(x(x+3)-1(x+3))\\=3(x+3)(x-1)[/tex]
So, the sign function is,
Therefore, [tex]f'(x)>0[/tex] when [tex]x\in (-\infty ,-3)\cup (1,\infty )[/tex]
So, [tex]f(x)[/tex] is increasing on [tex]x<-3[/tex] or [tex]x>1[/tex]
Hence, the correct option is an option (c).
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