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The function $f(x)$ defined as $f(x)=x^3-6x^2+24x$, where $x$ is real, is

  1. strictly increasing
  2. strictly decreasing
  3. increasing in $(- \infty, 0)$ and decreasing in $(0, \infty)$
  4. decreasing in $(- \infty, 0)$ and increasing in $(0, \infty)$ 
in Calculus
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Is $A$ the answer?
1

Answer $$

Given: $$f(x) = x^3-6x^2+24x$$

First derivative of above equation is:

$$f'(x) = 3x^2-12x+24 = 3\underbrace{(x^2-4x+8)}_\text{D<0}$$

Since, $$a > 0, D = b^2 -4ac < 0 \implies$$ Parabola is upward facing with no real roots.

 

Now, Vertex of the parabola, $$x-coordinate = \frac{-b}{2a} = 2$$

Putting this value in above equation, we will get the y-coordinate:

$$y-coordinate = 3(2^2-4.2+8) = 12 $$

So, the coordinates of the vertex are $$(2, 12)$$

0

@ankitgupta.1729

@srestha

Can you please verify this answer.

1

@`JEET

according to me, it is correct.

1
Thanks

1 Answer

1 vote

Answer $A$

Given: $$f(x) = x^3-6x^2+24x$$

First derivative of above equation is:

$$f'(x) = 3x^2-12x+24 = 3\underbrace{(x^2-4x+8)}_\text{D<0}$$

Since, $$a > 0, D = b^2 -4ac < 0 \implies$$ Parabola is upward facing with no real roots.

 

Now, Vertex of the parabola, $$x-coordinate = \frac{-b}{2a} = 2$$

Putting this value in above equation, we will get the y-coordinate:

$$y-coordinate = 3(2^2-4.2+8) = 12 $$

So, the coordinates of the vertex are $$(2, 12)$$

So, $(A)$ is the right option.


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