Answer:
Minimum 8 at x=0, Maximum value: 24 at x=4
Step-by-step explanation:
Retrieving data from the original question:
![f(x)=x^{2}+8\:over\:[-1,4]](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E%7B2%7D%2B8%5C%3Aover%5C%3A%5B-1%2C4%5D)
1) Calculating the first derivative
![f'(x)=2x](https://tex.z-dn.net/?f=f%27%28x%29%3D2x)
2) Now, let's work to find the critical points
Set this
0, belongs to the interval. Plug it in the original function
![f(0)=(0)^2+8\\f(0)=8](https://tex.z-dn.net/?f=f%280%29%3D%280%29%5E2%2B8%5C%5Cf%280%29%3D8)
3) Making a table x, f(x) then compare
x| f(x)
-1 | f(-1)=9
0 | f(0)=8 Minimum
4 | f(4)=24 Maximum
4) The absolute maximum value is 24 at x=4 and the absolute minimum value is 8 at x=0.
Answer:
B
Step-by-step explanation:
I am pretty sure your answer is this because 3+2=5 and 5*7=35 abc
We know that the slope formula is m = (y1 - y2)/(x1 - x2)
Your points are: (3, 7) and (4, -8)
In this case,
x1 = 3
x2 = 4
y1 = 7
y2 = -8
Now, just plug in the numbers:
m = (7 - -8)/(3-4)
m = 15/-1
m = -15
Your slope is -15.