Question

If f (x) = x^3 – 3x^2, find the values of x for which f’(x) = 0.

Answer 1

f(x) = g(x)

⇒ 3x2-l=3+x ⇒ 3x2-x-4 = 0 ⇒ (3x – 4)(x+ 1) – 0  

x= -1,4/3

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Answer 2

The values of x is either 0 or 2.

Step-by-step explanation:

According to the given information, the given polynomial function is f (x) = x^3 – 3x^2. We need to find the values of x when f’(x) = 0.

Now, differentiating f(x) = x^3 – 3x^2 with respect to x, we get, by using the rule of differentiation that is $\frac{d}{dx} (x^{n} )=nx^{n-1}$,

$\frac{d}{dx} (x^{3} -3x^{2} )\\=\frac{d}{dx} (x^{3} )-\frac{d}{dx}(3x^{2})\\=3x^{2} -3(2x)\\=3x^{2} -6x$

Now differentiation of f(x) is equal to f'(x). According to the given information, f'(x)=0.

Then, putting the value in the above equation, we get,

$3x^{2} -6x=0$

Taking x common from both the terms we get,

$x( 3x -6) = 0$

Then either x = 0 or 3x - 6 = 0.

Now, 3x-6 = 0 that is, 3x = 6 that is, x = 2.

Then, the values of x is either 0 or 2.

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