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$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$ $($ true or false$)?$

Here $f(0) = -46$ and $f(6) = 2$ since function is continuous so it must have at least one root between $0$ and $6$, but how to check if it has all its roots between $0$ and $6$, without really finding out the roots?

Mathaddict
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  • As the question asks whether or not all real roots are between $6$ and $7$, finding even one real root outside of that range, as you have done, shows whether or not the requested statement is true. You don't need to show all roots are between $0$ and $6$ or any other such range. Or is the issue possibly due to a typo and it's to check for all real roots are between $0$ and $6$ instead? – John Omielan Jan 22 '19 at 03:43
  • We can check that the $f'(x)>0$ for all $x<0$ as well as $x>6$. Since $f(0)<0$, $f(x)<0$ for $x<0$. Similarly, $f(x)>0$ for $x>6$. – Kelvin Soh Jan 22 '19 at 03:43
  • Hint: what is $f(2)$? – J. W. Tanner Jan 22 '19 at 03:45
  • $f(2)=2$ and also $f(5)=-1$ then we are done. – Offlaw Jan 22 '19 at 03:46
  • I'm really sorry . I edited the question, it is to find if f has all its roots between 0 and 6 or not. – Mathaddict Jan 22 '19 at 03:51
  • Another approach is to use the relatively straightforward, although a bit convoluted, method to directly determine the roots of a cubic equation (e.g., as shown in Cubic function), so you can confirm where the real roots are. Note, however, this is quite possibly not the simplest & easiest method you can use. Also, it does not work very well for higher power polynomials, so it's not a particularly general way to solve this type of problem. – John Omielan Jan 22 '19 at 03:58

4 Answers4

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Check $$f(1),f(2)$$ and $$f(4),f(5)$$ and $$f(5),f(6)$$

Dr. Sonnhard Graubner
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  • $f(1)=-13, f(2) = 2, f(4) = 2, f(5) = -1, f(6)=2$ so, $f$ has three real roots between 0 and 6. And f has almost 3 real roots, so all roots are in between 0 and 6. How would I estimate these points, where signs of function are alternatively positive and negative to solve such problems? – Mathaddict Jan 22 '19 at 04:19
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    @Mathsaddict: without the $+2$ the roots are at $2,4,6$. The big worry is that the root at $6$ will shift outside the interval. As the graph is rising through $6$ on its way to $+\infty$ lifting it up by $2$ will move the root to the left. Points halfway between the old roots are good candidates to not switch signs, so evaluating at $1,2,3,4,5,6$ is a good way to go. – Ross Millikan Jan 22 '19 at 04:27
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Well,

$$\alpha >0 \to f(-\alpha)=2-(\alpha+2)(\alpha+4)(\alpha+6) < -46$$ and

$$f(6+\alpha)=2+(4+\alpha)(2+\alpha)(\alpha)> 2$$ So at the very least all its real roots are $\in (0,6)$ You just need to show all of its roots are real.

Rhys Hughes
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    The statement "has all real roots between $0$ and $6$" can be interpreted in $2$ ways. One is that all of the roots are real and in that range, and the other is that any roots which are real (e.g., perhaps there is only $1$ real root) are in that range. When I first read that statement, I assumed the latter case. I believe the question could be worded better to avoid this potential ambiguity. – John Omielan Jan 22 '19 at 04:23
  • @JohnOmielan you misquoted me, and thusly caused your own confusion. I said all of the function's real roots were in $(0,6)$, not that the function has all real roots in $(0,6)$ – Rhys Hughes Jan 22 '19 at 04:42
  • Actually, I was quoting from the question, not what you wrote. My interpretation was that you just had to show that any roots which are real are in $\left(0,6\right)$, which you showed in the first part of your answer. However, if you interpret the question as stating that all roots are real and are in the $\left(0,6\right)$ range, then your last sentence makes sense. It's a relatively minor point which I perhaps shouldn't have even mentioned so, if you wish, please ignore this and I can delete my comments. – John Omielan Jan 22 '19 at 04:48
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In this problem, you don't have to actually calculate the roots (as you pointed out). One way you could solve this is to use the intermediate value theorem (going off of Dr. Sonnhard Graubner's comment):

f(1) = -13

f(2) = 2

By the intermediate value theorem, you know that f has to pass through 0 at some point between f(1) and f(2), meaning that one of its roots is between x=1 and x=2, fulfilling the requirement.

f(4) = 2

f(5) = -1 Same process as above - by the intermediate value theorem, you know there must be another root between x=4 and x=5.

f(6) = 2 You also know there must be a root between x=5 and x=6.

elbecker
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Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$

1)$F(x), f(x)$ polynomials of degree $3.$

2) $x \rightarrow \infty$ $F(x), f(x) \rightarrow \infty.$

3) $x \rightarrow -\infty$ $F(x), f(x) \rightarrow -\infty.$

4) Roots of $F(x)$ at $x=2,4,6$. That's it.

5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.

No roots of $f$ for $x>6$.

6) For $x <2$, $F(x) <0$, and

$F(x)= -(2-x)(4-x)(6-x)$ is strictly decreasing.

7)Likewise for $x <2$ $f(x)= F(x)+2$ is strictly decreasing

$F(1)=-(1)(3)(5)=-15$, and $f(1)= -15+2=-13$.

No root of $f$ for $x <1$.

Hence?

Peter Szilas
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