Given:
The function is
![f(x)=-x^2+4](https://img.qammunity.org/2021/formulas/mathematics/high-school/nq453d9k51r5k1mqb25g6pm1qi9kwzqc4a.png)
It defined on the interval -8 ≤ x ≤ 8.
To find:
The intervals on which the function is increasing and the interval on which decreasing.
Explanation:
We have,
![f(x)=-x^2+4](https://img.qammunity.org/2021/formulas/mathematics/high-school/nq453d9k51r5k1mqb25g6pm1qi9kwzqc4a.png)
Differentiate with respect to x.
![f'(x)=-(2x)+(0)](https://img.qammunity.org/2021/formulas/mathematics/high-school/iwjqbmzipc2vxsjuo47zpwfp2jc1c28q13.png)
![f'(x)=-2x](https://img.qammunity.org/2021/formulas/mathematics/high-school/xs6co17obt1cvyzldczica878xvoxyvi71.png)
For turning point f'(x)=0.
![-2x=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/reevlb7ygxkgg60ze65flikwe0k79yyo1j.png)
![x=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/6enp4x8l6ye502n229t5aopx507rpkpsln.png)
Now, 0 divides the interval -8 ≤ x ≤ 8 in two parts [-8,0] and [0,8]
For interval [-8,0], f'(x)>0, it means increasing.
For interval [0,8], f'(x)<0, it means decreasing.
Therefore, the function is increasing on the interval [-8,0] and decreasing on the interval [0,8].