The digit to the right of 1.249 is
1 answer:
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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
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.