Use the T-table below to determine 3 solutions to this equation: y = 2x
1 answer:
The table is attached as a figure
The given equation is ⇒⇒⇒⇒ y = 2x
To solve this equation, we need to pick numbers from the table then this number will be substituted into the equation to find y
we need 3 solutions . so, we need to pick 3 numbers of x
From the table, let us choose x = 0
y = 2x
y = 2 * (0)
y = 0
From the table, let choose second value of x such as x = 1
y = 2 * (1)
y = 2
From the table, let choose third value of x such as x = -1
y = 2 * (-1)
y = -2
So, the picked three solutions are
y = 0 at x = 0
y = -2 at x = -1
y = 2 at x = 1
The given equation is ⇒⇒⇒⇒ y = 2x
To solve this equation, we need to pick numbers from the table then this number will be substituted into the equation to find y
we need 3 solutions . so, we need to pick 3 numbers of x
From the table, let us choose x = 0
y = 2x
y = 2 * (0)
y = 0
From the table, let choose second value of x such as x = 1
y = 2 * (1)
y = 2
From the table, let choose third value of x such as x = -1
y = 2 * (-1)
y = -2
So, the picked three solutions are
y = 0 at x = 0
y = -2 at x = -1
y = 2 at x = 1
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Usually one will differentiate the function to find the minimum/maximum point, but in this case differentiating yields:
which contains multiple solution if one tries to solve for x when the differentiated form is 0.
I would, though, venture a guess that the minimum value would be (approaching) 5, since the function would be undefined in the vicinity.
If, however, the function is
Then differentiating and equating to 0 yields:
which gives:
or 
We reject x=5 as it is when it ix the maximum and thus,
, for 
which contains multiple solution if one tries to solve for x when the differentiated form is 0.
I would, though, venture a guess that the minimum value would be (approaching) 5, since the function would be undefined in the vicinity.
If, however, the function is
Then differentiating and equating to 0 yields:
which gives:
We reject x=5 as it is when it ix the maximum and thus,