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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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0
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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