Question

Suppose that ff is a differentiable function on the real line, we have −2≤f′(x)≤4−2≤f′(x)≤4 for all real numbers xx and f(2)=4f(2)=4. The largest possible value for f(7)f(7) is . The smallest possible value for f(7)f(7) is . The largest possible value for f(−2)f(−2) is . The smallest possible value for f(−2)f(−2) is

Answer 1

Answer:

(a)-6 and 24

(b)-12 and 12.

Step-by-step explanation:

Using the mean value theorem,

Suppose that f is a function which is continuous on a closed interval [a, b] and differentiable on an open interval (a, b), then there

exists a number c in (a, b) so that

$f^{'}(c)= \frac{f(b)-f(a)}{b-a}$

for every closed interval [a,b].

−2≤f′(x)≤4 and f(2)=4

First, we determine the largest and smallest possible value for f(7).

In the Interval [2,7],

$f^{'}(x)= \frac{f(7)-f(2)}{7-2}$

Since −2≤f′(x)≤4

$-2 \leq \frac{f(7)-f(2)}{5} \leq 4$

Recall that: f(2)=4

$-2 \leq \frac{f(7)-4}{5} \leq 4$

$-2X5 \leq f(7)-4 \leq 4X5$

$-10+4 \leq f(7)-4+4 \leq 20+4$

$-6 \leq f(7) \leq 24$

The greatest and least value are 24 and -6 respectively.

Similarly for f(-2)

In the Interval [-2,2],

$f^{'}(x)= \frac{f(2)-f(-2)}{2-(-2)}$

Since −2≤f′(x)≤4

$-2 \leq \frac{f(2)-f(-2)}{4} \leq 4$

Recall that: f(2)=4

$-2 \leq \frac{4-f(-2)}{4} \leq 4$

$-2X4 \leq -f(-2)+4 \leq 4X4$

$-8-4 \leq -f(-2)-4+4 \leq 16-4$

$-12 \leq -f(-2) \leq 12$

Dividing all through by negative

$-12 \leq f(-2) \leq 12$

The greatest and least value are 12 and -12 respectively.