In division, exponents with the same base are subtracted.
example:
$\frac{x^4}{x^2} = x^2 \\ \\ \frac{ x^{3} y^{2} }{ x^{6} y} = \frac{y}{ x^{3} } \\ \\ Negative-exponents \\ \\ x^{-2} = \frac{1}{ x^{2} }$

5 0

2 years ago

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

Shtirlitz [24]

Answer:

The number c is 2.

Step-by-step explanation:

Mean Value Theorem:

If f is a continuous function in a bounded interval [0,4], there is at least one value of c in (a,b) for which:

$f(c) = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx$

In this problem, we have that:

$f(x) = x, a = 0, b = 4$

So $f(c) = c$

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$f(c) = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx$

$c = \frac{1}{4-0}\int\limits^0_4 {x} \, dx$

$c = \frac{1}{4-0}*8$

$c = 2$

The number c is 2.

3 0

2 years ago