720 pt/h= ? qt/min please show step by step
1 answer:
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Answers:
a = 2 and b = -4
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Explanation:
Let's define the three helper functions
- f(x) = ax^2 - b
- h(x) = 6
- j(x) = 5ax+b
which are drawn from the piecewise function. The g(x) function will change depending on what the input is.
- If x < 1, then g(x) = f(x).
- If x = 1, then g(x) = h(x)
- If x > 1, then g(x) = j(x)
Since we want g(x) to be continuous at x = 1, this must mean the three functions f(x), h(x), j(x) must have the same output value when the input is x = 1.
Because h(x) = 6 is a constant function, the output is always 6 regardless of the input. Therefore, we want f(x) and j(x) to have 6 as their output when x = 1. Or else, the pieces won't connect.
Plug x = 1 into the f(x) function to get
f(x) = ax^2 - b
f(1) = a(1)^2 - b
f(1) = a - b
Set this equal to the desired output of 6 and we end up with the equation a-b = 6. Solving for 'a' leads to a = b+6.
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We'll use the same idea for j(x)
j(x) = 5ax + b
j(1) = 5a(1) + b
j(1) = 5a + b
5a+b = 6
5(b+6) + b = 6 ... plug in a = b+6; solve for b
5b+30+b = 6
6b+30 = 6
6b = 6-30
6b = -24
b = -24/6
b = -4
Which then leads to,
a = b+6
a = -4+6
a = 2
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Since a = 2 and b = -4, we go from this
to this
Meaning
f(x) = 2x^2+4 and j(x) = 10x-4
You should find that plugging x = 1 into each of those two functions leads to 6 as the output.
The graph is shown below. Note the red graph f(x) is only drawn when x < 1. Similarly, j(x) is only drawn when x > 1. The orange point represents h(x) which only happens when x = 1. So as the name implies, the piecewise function g(x) is composed of pieces of the three functions f(x), h(x), j(x).
