A function is defined as f(x) = 2x + 3 + 64/x^2. Find the values of x for which the function is increasing. So, I differentiated it and I get f'(x) as 2 - 128/x^3. For the function to be increasing the gradient must be positive. so 2 - 128/x^3 0. Solving this I can get x 4. However, it's obvious that it also works when x

Question

A function is defined as f(x) = 2x + 3 + 64/x^2. Find the values of x for which the function is increasing.

So, I differentiated it and I get f'(x) as 2 - 128/x^3. For the function to be increasing the gradient must be positive. so 2 - 128/x^3 > 0.

Solving this I can get x > 4. However, it's obvious that it also works when x < 0. (Taking -1, -2 -  (-128/1) = -2 + 128 = 126. Which is greater than 0.

What am I doing that makes me lose an answer?

Thanks.

anonymous · Answer

2 + 128 = 130, which is greater than 0. I put an extra minus in there but the problem's the same.

anonymous · Answer

is it all over x^2?

anonymous · Answer

$$f(x)=2x+3+\frac{64}{x^2}$$?

anonymous · Answer

Just the 64 over the x^2.

anonymous · Answer

if so you get 
$$f'(x)=2-\frac{128}{x^3}=\frac{2(x^3-64)}{x^3}$$ yes?

anonymous · Answer

positive if 
$$x<0$$ or $$x>4$$

anonymous · Answer

oh i see i should have read your question more carefully. yes you cannot ignore the denominator.  it counts too

anonymous · Answer

you get 
$$\frac{2(x-4)(x^2+4x+8)}{x^3}$$

anonymous · Answer

so critical points at 4 and 0 
it is negative between the critical points and positive outside them

anonymous · Answer

you cannot just look at the factor x - 4  you also  have to consider x^3 as a factor, which is positive if x is and negative if x is negative

anonymous · Answer

this is clear yes?

anonymous · Answer

@alchemista i am working with the derivative

anonymous · Answer

Where did you get 
2(x−4)(x2+4x+8)\x3
from?

anonymous · Answer

your derivative is 
$$f'(x)=2-\frac{128}{x^3}=\frac{2x^3-128}{x^3}=\frac{2(x^3-64)}{x^3}$$ yes?

anonymous · Answer

in order to see where this is positive and negative we need to write over one denominator

anonymous · Answer

Yes.

I did 

2- 128/x^3 > 0

2x^3 - 128 > 0

x^3 > 64

anonymous · Answer

then i just factored 
$$x^3-64$$ as difference of two cubes

anonymous · Answer

oh no 

$$2-\frac{128}{x^3}>0$$ not the same as 
$$2x^3-128>0$$

anonymous · Answer

because 
$$2-\frac{128}{x^3}\neq 2x^3-128$$

anonymous · Answer

you have to make the fraction first  and write 
$$\frac{2x^3-128}{x^3}>0$$ and consider both numerator and denominator

anonymous · Answer

clear or no?

anonymous · Answer

Ok, so where do you go from there?