MHB Calculating g(x) for Y=f(x) Passing Through Points

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The discussion focuses on calculating the function g(x) defined as g(x)=2f(x-1) using the points provided for f(x). The calculations for g(0), g(1), g(2), and g(3) are based on the values of f at specific points. For g(0), it is determined that g(0)=6 by evaluating f(-1), which equals 3. Participants are encouraged to continue with the calculations for g(1), g(2), and g(3). The interaction highlights collaborative problem-solving in mathematics.
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Y=f(x)
which passes through points:
(-1,3) and (0,2) and (1,0) and (2,1) and (3,5)
second function is defined: g(x)=2f(x-1)

Calculate g(0)
Calculate g(1)
Calculate g(2)
Calculate g(3)
 
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AndreArgo said:
Y=f(x)
which passes through points:
(-1,3) and (0,2) and (1,0) and (2,1) and (3,5)
second function is defined: g(x)=2f(x-1)

Calculate g(0)
Calculate g(1)
Calculate g(2)
Calculate g(3)

Hello, and welcome to MHB! (Wave)

We are given:

$$g(x)=2f(x-1)$$

We are also given 5 points in the form:

$$(x,f(x))$$

And so:

$$g(0)=2f(0-1)=2f(-1)=2\cdot3=6$$

Does that make sense? Can you continue?
 
MarkFL said:
Hello, and welcome to MHB! (Wave)

We are given:

$$g(x)=2f(x-1)$$

We are also given 5 points in the form:

$$(x,f(x))$$

And so:

$$g(0)=2f(0-1)=2f(-1)=2\cdot3=6$$

Does that make sense? Can you continue?
Just figured it out. Thank you Mark
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

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