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.

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.