Domain Of K In $f(x)=x^3+kx^2+5x+4\sin^2x - Math Stack Exchange

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Learn more about Teams Domain of k in $f(x)=x^3+kx^2+5x+4\sin^2x$ Ask Question Asked 6 years, 10 months ago Modified 6 years, 10 months ago Viewed 301 times 4 $\begingroup$

The question says:

Let $f(x)=x^3+kx^2+5x+4\sin^2x$ be an increasing function on $x \in R$. Then domain of k is?

This is What I tried:

I tried differentiating the expression but $f'(x)$ didn't turn out to be a quadratic(as it has that $\sin2x$ term) $$f'(x)=3x^2+2kx+5+4\sin2x>0$$

So how else am I supposed to solve this question?

• calculus
• derivatives

Share Cite Follow edited Feb 26, 2018 at 1:19 user99914 asked Feb 24, 2018 at 12:25 CarrickCarrick 19799 bronze badges $\endgroup$ 5

• $\begingroup$ @Netchaiev wolframalpha.com/input/?i=derivative+sin%5Bx%5D%5E2 $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Commented Feb 24, 2018 at 12:35
• $\begingroup$ @Netchaiev I know its $2\sin x \cos x$ but aren't they both the same?? $\endgroup$ – Carrick Commented Feb 24, 2018 at 12:35
• $\begingroup$ yes you are both right, my bad :) (deleting the first comment) $\endgroup$ – Netchaiev Commented Feb 24, 2018 at 12:37
• $\begingroup$ @Netchaiev not a big deal..its ok $\endgroup$ – Carrick Commented Feb 24, 2018 at 12:38
• $\begingroup$ @Netchaiev It's alright. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Commented Feb 24, 2018 at 12:39

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2 Answers 2

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 2 $\begingroup$

The set up of the problem is completely right with only one remark: you can set $$f'(x)\ge0$$ to have $f(x)$ increasing and maybe also strictly increasing since the the set on which $f'(x)=0$ for "regular functions" like the given doesn't contain any intervals.

You can refer also to this other answer for a detailed comment on this latter fact.

For a first estimation of the range note that for

$$3x^2+2kx+5\ge4$$

the condition is satisfy, thus from here we can find a first range for $k\in[-\sqrt 3,\sqrt 3]$, indeed

$$3x^2+2kx+1\ge0 \iff \Delta=4k^2-12\le 0$$

To improve the result we could try to find the minimum of $f'(x)$ and set it to $0$, that is

$$f''(x)=6x+2k+8\cos2x=0 \implies x=x_k$$

and

$$f'(x_k)=0$$

Share Cite Follow edited Feb 24, 2018 at 13:24 answered Feb 24, 2018 at 12:39 useruser 160k1313 gold badges8484 silver badges154154 bronze badges $\endgroup$ 8

• 1 $\begingroup$ I thought about that but won't it be wrong because we are fixing the value of x so that sin2x attains maxima or minima?? $\endgroup$ – Carrick Commented Feb 24, 2018 at 12:43
• $\begingroup$ It is a first good estimation since $4\sin2x\ge -4$ if $3x^2+2kx+5\ge4$ we are sure that $f'(x)\ge 0$. . $\endgroup$ – user Commented Feb 24, 2018 at 12:45
• $\begingroup$ sorry...didn't really get it...how does that answer my question?? $\endgroup$ – Carrick Commented Feb 24, 2018 at 12:51
• $\begingroup$ @Carrick No it doesn't answer it is only an hint, I'm thinking about it to improve the estimation and some one else will give other answers, wait to accept as solved, but you can upvote if you think it was useful to you. $\endgroup$ – user Commented Feb 24, 2018 at 12:53
• $\begingroup$ by question I didn't mean the actual queston..I was talking about what I just asked in the comments(fixing the value of x thing) and your hint is correct..it matches with the actual answer $\endgroup$ – Carrick Commented Feb 24, 2018 at 12:59

| Show 3 more comments 0 $\begingroup$

From $f '(x)=3x^2+2kx+5+4\sin2x$, try to solve $f'(x)=0$, so $$k=\frac{-3x^2-5-4\sin2x}{2x}$$ If you can't solve $f'(x)=0$, then $f(x)$ is increasing, so the domain of $k$ is everything outside the range of that function.

Share Cite Follow answered Feb 26, 2018 at 2:37 Empy2Empy2 51.9k11 gold badge4545 silver badges9595 bronze badges $\endgroup$ Add a comment |

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