What is the range of F(x)=72x-x^2

Mathematics

1 answer:

Marianna [84]2 years ago

3 0

Answer:

Range of F(x) = (-∞ ,72).

Step-by-step explanation:

Range of the function F(x) means the set of values which are taken by the function along its domain.

F(x) = 72 - $x^{2}$

Since coefficient of $x^{2}$ is negative on the right side ,

Maximum value of F(x) will come at x = 0.

Which is, F(0) = 72.

Now, as x increases , F(x) decreases , when x will reach infinity ,

F(x) will be reaching negative of infinity .

Thus, F(x) is ranging from -∞ to 72 .

Range of F(x) = (-∞ ,72).

You might be interested in

Find the average value of f over region

yan [13]

The area of D is given by:

$\int\limits \int\limits {1} \, dA = \int\limits_0^7 \int\limits_0^{x^2} {1} \, dydx \\ \\ = \int\limits^7_0 {x^2} \, dx =\left. \frac{x^3}{3} \right|_0^7= \frac{343}{3}$

The average value of f over D is given by:

$\frac{1}{ \frac{343}{3} } \int\limits^7_0 \int\limits^{x^2}_0 {4x\sin(y)} \, dydx = -\frac{3}{343} \int\limits^7_0 {4x\cos(x^2)} \, dx \\ \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49$