Question

if f(x)=x(x+1)(x+2)(x+3),then show that f'(x) has three real roots​

Answer 1

Step-by-step explanation:

three roots -1,-2,-3

hope this answer

Answer 2

Given:

f (x) = $x (x + 1) (x + 2) (x + 3).$

To Find:

We have to show that if f (x) = x (x + 1) (x + 2) (x + 3), then f'(x) has 3 real roots.

Solution:

Given that f (x) = $x (x + 1) (x + 2) (x + 3)$.

Opening the brackets we get,

f (x) = $x^{4} + 5x^{3} + 9x^{2} + 9x.$

f'(x) = $\frac{d}{dx} (x)$

∴, f'(x) = $\frac{d}{dx} (x^{4} + 5x^{3} + 9x^{2} + 9x)$

f'(x) = $4.x^{3} + 5.3.x^{2} + 9.2.x^{1} + 9.1 = 4x^{3} + 15x^{2} + 18x + 9.$

Put x = 0, in above equation.

f'(x) = $4 .0^{3} + 15.0^{2} + 18.0 + 9 = 0.$

Put x = -1 to get,

f'(x) = $4.(-1)^{3} + 15.(-1)^{2} + 18.(-1) + 9 = 0.$

Put x = -3 to get,

f'(x) = $4.(-3)^{3} + 15.(-3)^{2} + 18.(-3) + 9 = 0.$

On solving the above cubic polynomial, we get three real roots.

The positive root = 0

The negative roots = 3 and 1.

Hence the real roots of f'(x) are 0, -1, and -3.

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