3 + p = 8
Subtract 3 on both sides
3 + p - 3 = 8 - 3
Simplify both sides
p = 5
Subtract 3 on both sides
3 + p - 3 = 8 - 3
Simplify both sides
p = 5
Solve the following equation algebraically: x^2=20
2 answers:
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<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>
Answer:
1) There is no maximum height
2) The ball will never land
Step-by-step explanation:
<u>Derivatives</u>
Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.
Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that
If f''(a) is positive, then x=a is a minimum
If f''(a) is negative, then x=a is a maximum
1)
The function provided in the question is
Let's find the first derivative
solving h'=0:
x=3
Computing h''
h''(x)=4
It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height
2)
To find when will the ball land, we set h=0
Simplifying by 2
Completing squares
Factoring and rearranging
There is no real value of x to solve the above equation, so the ball will never land.
This problem seems to be wrong because no minimum point was found and no point of landing exists