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Is the set of all differentiable functions whose derivative at $t=1$ is three, a vector space?

I know how to show the set of functions is a vector space, but why is this not a vector space. I know how to check if it is closed under scalar multiplication and vector addition, but I'm not sure how the derivative is put it to play. I would like to know how to show it is not a vector space.

Thank you.

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    $\begingroup$ Is the sum of 2 such functions still in the set? $f(x)+g(x)$ is such that $f'(3)+g'(3)=...$ $\endgroup$ – Bernard Masse Oct 27 '16 at 20:50
  • $\begingroup$ @BernardMasse For this problem though t=1, so wouldn't f′(1)+g′(1)=3+3, which is not a function whose derivative at t=1 is 3... Is this what I take away? $\endgroup$ – Struggling_College_Student Oct 27 '16 at 20:54
  • $\begingroup$ "Differentiable," not "differential." $\endgroup$ – symplectomorphic Oct 27 '16 at 21:41
  • $\begingroup$ If $f$ is in the set, is $(1/2)f$ in the set? $\endgroup$ – zhw. Nov 1 '16 at 23:20
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A set $\mathbb{V}$ is a vector space if and only if, taking two elements $v$ and $w$ belonging to $\mathbb{V}$, and given any two real numbers $\alpha, \beta$ then $\alpha v+\beta w \in \mathbb{V}$. In your case, taken two differentiable functions $f(t)$ and $g(t)$ whose derivative is equal to 3 when $t=1$ you can see that their sum is not meeting this condition anymore, since, calling $h(t):=f(t)+g(t)$: $$h'(1) = (f(t)+(g(t))'|_{t=1}=(f'(t)+g'(t))|_{t=1} = f'(1)+g'(1)=3+3=6\ne3$$ In the end, this set is not a vector space.

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Hint: Take for example $f(t)=t^{3}$ and $g(t)=t^{3}+2$

$f(t)+g(t)=t^{3}+t^{3}+2$

Now take the derivative

Is it equal to $3$ at $t=1$ ?

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