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Problem: Suppose ∃ function h: ℝ → ℝ such that h as a second derivative h''(x) = 0 ∀ x ∈ ℝ. Prove ∃ numbers m, b:

h(x) = mx + b, ∀ x ∈ ℝ.


My attempt: Consider h(x) = mx + b, with constants m and b. Note:

h(x) = mx + b

h'(x) = m + 0

h''(x) = 0.

Hence the statement.


This feels a little weak to me. I'd appreciate any suggestions on how better to prove (or if I'm totally off base, maybe set me on the right track). Thanks for any help in advance.

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  • $\begingroup$ You need to show that if $h''(x)$ then $h(x) = mx+b$, not that if $h(x) =mx+b$ then $h''(x)=0$. You can be done by integrating $h''(x) = 0$ two times. $\endgroup$
    – Winther
    Commented Mar 18, 2015 at 0:29
  • $\begingroup$ The reason that you're proof does not work is that, at the beginning, you say, "Consider $h(x) = mx + b$, with constants $m$ and $b$." You're assuming that such constants exist, but that is what you are trying to prove. So you've proved that $h(x) = mx + b \Rightarrow h''(x) = 0$, while the question asks you to prove that other direction. $\endgroup$
    – Mark
    Commented Mar 18, 2015 at 0:37

3 Answers 3

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I am also not sure, but i would go the other way: $h''(x) = 0 \Rightarrow \int h''(x) dx = h'(x)= m \Rightarrow \int h'(x) dx = h(x) =mx + b$

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Because integration is the inverse of differentiation, we can find that if $h(x)=0,$

$$\int h ''(x) \,dx=\int 0\, dx\implies h'(x)=C_1\implies \int h'(x) \,dx=\int C_1 \,dx\implies h(x)=xC_1+C_2$$

Now we can choose $m=C_1$ and $b=C_2$, and we are done.

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Solution 1 (with no integration). If $x\geq 0$, by the Mean Value Theorem (MVT) there exists $c_1>0$ such that $$h(x)-h(0)=h'(c_1)x.\tag{1}$$

If $x<0$, again by the MVT, there exists $c_2>0$ such that $$h(0)-h(x)=h'(c_2)(-x).\tag{2}$$

But, applying the MVT again, we conclude that $$h'(c_1)-h'(c_2)=0(c_1-c_2)=0$$ and thus $h'(c_1)=h'(c_2)$. It follows from $(1)$ and $(2)$ that there exists $m=h'(c_1)$ and $b=h(0)$ such that, for any $x\in\mathbb{R}$, $$h(x)=mx+b.$$

Solution 2 (with integration). By the Fundamental Theorem of Calculus (FTC),

$$h'(t)=h'(0)+\int_0^th''(x)\;dx=h'(0)+\int_0^t0\;dx=h'(0)+0=h'(0), \quad\forall\ t\in\mathbb{R}.$$

Take $x\in\mathbb{R}$. By the FTC again,

$$h(x)=h(0)+\int_0^x h'(t)\;dt=h(0)+\int_0^x h'(0)\;dt=h(0)+h'(0)\int_0^x 1\;dt=h(0)+h'(0)x$$

So, there exists $m=h'(0)$ and $b=h(0)$ such that $h(x)=mx+b$.

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