Let the function $$f : [-7, 0] \rightarrow R$$ be continuous on $$[-7, 0]$$ and differentiable on $$(-7, 0)$$. If $$f(-7) = -3$$ and $$f'(x) \le 2$$ for all $$x \in (-7, 0)$$, then for all such functions $$f$$, $$f(-1) + f(0)$$ lies in the interval

We have a function $$f:[-7,0]\to\mathbb R$$ which is continuous on the closed interval $$[-7,0]$$ and differentiable on the open interval $$(-7,0)$$. The values already known are $$f(-7)=-3$$ and, for every $$x\in(-7,0)$$, the derivative satisfies $$f'(x)\le 2$$.

To relate the values of the function at different points we use the Mean Value Theorem (MVT). The MVT states that for any two points $$a,b$$ in the interval with $$a<b$$, there exists a point $$c\in(a,b)$$ such that

$$f(b)-f(a)=f'(c)\,(b-a).$$

Because $$f'(c)\le 2$$, the difference $$f(b)-f(a)$$ cannot exceed $$2(b-a)$$. Expressed as an inequality,

$$f(b)\le f(a)+2\,(b-a).$$

We now apply this inequality twice, once for the pair $$(-7,-1)$$ and once for the pair $$(-7,0)$$.

First take $$a=-7,\;b=-1$$. Then $$b-a=(-1)-(-7)=6$$, so

$$f(-1)\le f(-7)+2\cdot 6 =-3+12 =9.$$ Thus

$$f(-1)\le 9.$$

Next take $$a=-7,\;b=0$$. Now $$b-a=0-(-7)=7$$, hence

$$f(0)\le f(-7)+2\cdot 7 =-3+14 =11.$$ So

$$f(0)\le 11.$$

Adding the two individual inequalities gives an upper bound for the required sum:

$$f(-1)+f(0)\le 9+11=20.$$ Hence, whatever the admissible function,

$$f(-1)+f(0)\le 20.$$

We must also decide whether there is a finite lower bound. Observe that the derivative condition $$f'(x)\le 2$$ places no restriction on how negative the derivative may be; it can be any negative number. Therefore the function is free to decrease as steeply as we like.

To see this concretely, fix a very large positive number $$N$$ and construct a simple piecewise-linear example:

For $$x\in[-7,-1]$$ set $$f(x)=-3-N\,(x+7).$$ Here the constant slope is $$f'(x)=-N$$, which respects the inequality $$f'(x)\le 2$$ because $$-N<2$$ for every positive $$N$$.

At $$x=-1$$ the value is

$$f(-1)=-3-N\,(6)=-3-6N.$$

On the short interval $$[-1,0]$$, let the function rise with the maximum permitted slope $$2$$: $$f(x)=f(-1)+2\,(x+1),\qquad x\in[-1,0].$$ This keeps the derivative equal to $$2$$ and ensures differentiability at $$x=-1$$.

Then at $$x=0$$ we obtain

$$f(0)=f(-1)+2\,(0+1) =(-3-6N)+2 =-1-6N.$$

The sum of the two required values is therefore

$$f(-1)+f(0)=(-3-6N)+(-1-6N)=-4-12N.$$

As $$N\to\infty$$ the number $$-4-12N$$ tends to $$-\infty$$. So by choosing a sufficiently large $$N$$, the sum $$f(-1)+f(0)$$ can be made arbitrarily negative. Consequently, there is no finite lower bound; the set of all possible values extends indefinitely towards $$-\infty$$.

Combining the results, we have established that

$$-\infty<f(-1)+f(0)\le 20.$$

Thus the entire range of admissible sums is the interval $$(-\infty,\,20]$$.

Hence, the correct answer is Option A.