Question

Decide whether the function f=4−x^2 satisfies the hypotheses of the MVP on the interval [a,b]=[−1,0].
Then find all values of c in the interval [a,b] satisfying f′(c)=(f(b)−f(a))/b−a.
If there is more more than one enter them as a comma separated list.

c=________
Enter NONE if there are no such points in the interval

Answer 1

By MVP do you mean the mean value theorem?

Answer 2

And, all right, first: what is the derivative of the function?

Answer 3

@LolWolf give me a sec

Answer 4

the derivative is -2x

Answer 5

Yessir/ma'am, now find all of the values that satisfy:
\[
\frac{df}{dx}=-2x=\frac{f(b)-f(a)}{b-a}=\frac{f(0)-f(-1)}{0-(-1)}=?
\]

Answer 6

@LolWolf ok so the the interval (-1,0) will go to the 4-x^2? when it says satisfy does it mean to equal to the -2x?

Answer 7

Yep. Although, I'm not quite sure what you mean by the first part "the interval (-1,0) will go to the 4-x^2." If you mean that the interval's boundary is what will be used, then, yes.

Answer 8

what i meant to say is that do substitute the interval in the 4-x^2

Answer 9

The end points of the interval. Check my third reply, above---the interval's endpoints are \(a\) and \(b\), where \(b>a\), which is what you would use for this.

Answer 10

im still lost, like i substitute but i get it wrong,

Answer 11

@LolWolf

Answer 12

yes, it satisfies the hypothesis of the MVT.

Answer 13

Hum... let's actually work it out:

\[
\frac{df}{dx}=-2x=\frac{f(b)-f(a)}{b-a}=\frac{f(0)-f(-1)}{0-(-1)}=\frac{4-3}{1}=1
\]Hence:
\[
x=-\frac{1}{2}
\]

Answer 14

thank you

Answer 15

you can prove for yourself that is \(f\) is a quadratic polynomial, then the elusive \(c\) that satisfies the conclusion of the MVT sits right in the middle of the interval

Answer 16

in this case your interval was \([−1,0]\) and \(c=-\frac{1}{2}\)