Let the functionf : [-7, 0]→R be continuous on [–7, 0] and differentiable on (–7, 0). If f(–7) = –3 and ⨍ ’(x)≤ 2 for all x ∈(-7, 0) , then for all such function ⨍ , ⨍ (–1) + ⨍ (0) is in the interval:

f1(x)=f(-1)-f(-7)-1-(-7)≤2 =f(-1)+36≤2 ⇒f(-1)≤9 f1(x)=f(0)-f(-7)0+7≤2 =f(0)+3≤14 ⇒ f(0)≤11 f(-1)+f(0)≤20

Let the functionf : [-7, 0]→R be continuous on [–7, 0] and differentiable on (–7, 0). If f(–7) = –3 and ⨍ ’(x)≤ 2 for all x ∈(-7, 0) , then for all such function ⨍ , ⨍ (–1) + ⨍ (0) is in the interval:

f1(x)=f(-1)-f(-7)-1-(-7)≤2 =f(-1)+36≤2 ⇒f(-1)≤9 f1(x)=f(0)-f(-7)0+7≤2 =f(0)+3≤14 ⇒ f(0)≤11 f(-1)+f(0)≤20

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Let the function $f : [- 7, 0] \to R$ be continuous on [–7, 0] and differentiable on (–7, 0). If f(–7) = –3 and $⨍ ’ (x) \leq 2$ for all $x \in (- 7, 0),$ then for all such function $⨍, ⨍ (- 1) + ⨍ (0)$ is in the interval:

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$[- 3, 11)$

$(- \infty, 20]$

(–6, 20)

$(- \infty, 11]$

answer is B.

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$f^{1} (x) = \frac{f (- 1) - f (- 7)}{- 1 - (- 7)} \leq 2 = \frac{f (- 1) + 3}{6} \leq 2 \Rightarrow f (- 1) \leq 9 f^{1} (x) = \frac{f (0) - f (- 7)}{0 + 7} \leq 2 = f (0) + 3 \leq 14 \Rightarrow f (0) \leq 11 f (- 1) + f (0) \leq 20$

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Let the functionf : [-7, 0]→R be continuous on [–7, 0] and differentiable on (–7, 0). If f(–7) = –3 and ⨍ ’(x)≤ 2 for all x ∈(-7, 0) , then for all such function ⨍ , ⨍ (–1) + ⨍ (0) is in the interval:

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Let the function $f : [- 7, 0] \to R$ be continuous on [–7, 0] and differentiable on (–7, 0). If f(–7) = –3 and $⨍ ’ (x) \leq 2$ for all $x \in (- 7, 0),$ then for all such function $⨍, ⨍ (- 1) + ⨍ (0)$ is in the interval: