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

The question is an illustration of mean value theorem.

The given parameters are:

The mean value theorem states that:

Substitute 7 and 2 for b and a, respectively

Multiply both sides by 5

Substitute 4 for f(2)

Add 4 to both sides

So, we have:

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

Recall that:

Substitute -2 and 2 for b and a, respectively

Multiply both sides by -4

Substitute 4 for f(2)

Add 4 to both sides

So, we have:

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

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(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

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],

Since −2≤f′(x)≤4

Recall that: f(2)=4

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

Similarly for f(-2)

In the Interval [-2,2],

Since −2≤f′(x)≤4

Recall that: f(2)=4

Dividing all through by negative

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

12cm

the answer is d

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)