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

[TeX] f^{'}(c)= \frac{f(b)-f(a)}{b-a}[/TeX]

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],

[TeX] f^{'}(x)= \frac{f(7)-f(2)}{7-2}[/TeX]

Since −2≤f′(x)≤4

[TeX] -2$\leq$\frac{f(7)-f(2)}{5}$\leq$4[/TeX]

Recall that: f(2)=4

[TeX] -2$\leq$ \frac{f(7)-4}{5}$\leq$4[/TeX]

[TeX] -2X5$\leq$ f(7)-4$\leq$4X5[/TeX]

[TeX] -10+4$\leq$ f(7)-4+4$\leq$20+4[/TeX]

[TeX] -6$\leq$ f(7)$\leq$24[/TeX]

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

Similarly for f(-2)

In the Interval [-2,2],

[TeX] f^{'}(x)= \frac{f(2)-f(-2)}{2-(-2)}[/TeX]

Since −2≤f′(x)≤4

[TeX] -2$\leq$ \frac{f(2)-f(-2)}{4}$\leq$4[/TeX]

Recall that: f(2)=4

[TeX] -2$\leq$ \frac{4-f(-2)}{4}$\leq$4[/TeX]

[TeX] -2X4$\leq$ -f(-2)+4$\leq$4X4[/TeX]

[TeX] -8-4$\leq$ -f(-2)-4+4$\leq$16-4[/TeX]

[TeX] -12$\leq$ -f(-2)$\leq$12[/TeX]

Dividing all through by negative

[TeX] -12$\leq$ f(-2)$\leq$12[/TeX]

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

Answer 2

The question is an illustration of mean value theorem.

• The least and the greatest values of f(7) are -6 and 24, respectively.
• The least and the greatest values of f(-2) are -12 and 12, respectively.

The given parameters are:

$\mathbf{-2 \le f'(x) \le 4}$

$\mathbf{f(2) = 4}$

The mean value theorem states that:

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

Substitute 7 and 2 for b and a, respectively

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

$\mathbf{f^{'}(c)= \frac{f(7)-f(2)}{5}}$

Multiply both sides by 5

$\mathbf{5f^{'}(c)= f(7)-f(2)}$

Substitute 4 for f(2)

$\mathbf{5f^{'}(c)= f(7)-4}$

Add 4 to both sides

$\mathbf{4 + 5f^{'}(c)= f(7)}$

So, we have:

$\mathbf{4 + 5 \times -2 \le f(7) \le 4 + 5 \times 4}$

$\mathbf{-6 \le f(7) \le 24}$

Hence, the least and the greatest values of f(7) are -6 and 24, respectively.

Recall that:

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

Substitute -2 and 2 for b and a, respectively

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

$\mathbf{f^{'}(c)= \frac{f(-2)-f(2)}{-4}}$

Multiply both sides by -4

$\mathbf{-4f^{'}(c)= f(-2)-f(2)}$

Substitute 4 for f(2)

$\mathbf{-4f^{'}(c)= f(-2)-4}$

Add 4 to both sides

$\mathbf{4-4f^{'}(c)= f(-2)}$

So, we have:

$\mathbf{4 - 4 \times -2 \le f(-2) \le 4 - 4 \times 4}$

$\mathbf{12 \le f(-2) \le -12}$

Hence, the least and the greatest values of f(-2) are -12 and 12, respectively.

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