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Next we have to talk about evaluating functions. Evaluating a function is in fact nothing more than asking what its value is for particular values of x. Another way of looking at it is that we are asking what the y value is for a given x is.
Evaluation is actually quite simple. Let's consider the function we were looking at above
f( x ) = x2 - 5x + 3
and ask what its value is for x= 4 . In terms of function notation we will "ask" this using the notation f( 4) . Thus, while there is something other than the variable within the parenthesis we are actually asking what the value of the function is for that specific quantity.
Now, while we say the value of the function we are actually asking what the value of the equation is for that specific value of x. Here is f( 4) .
f ( 4)= ( 4)2 - 5 ( 4) + 3 = 16 - 20 +3 = -1
Notice that evaluating a function is done in exactly the same way in which we evaluate equations. We plug in for x whatever is on the inside of the parenthesis on the left. Following is another evaluation for this function.
f( -6) = ( -6)2 - 5 ( -6) + 3 = 36 + 30 + 3 =69
Thus, again, whatever is on the inside of the parenthesis on the left is plugged in for x in the equation on the right.
log6 X + log6 (x-5) = 1
The following relation is not a function. {(6,10) ( -7, 3) (0, 4) (6, -4)} Solution Don't worry regarding where this relation came from. It is only on
Draw the graph of y=x^2-4x from x=-1 to x=5.use the scale of 2cm on the x axis and 1cm on the y axis.Estimate the gradient at point:x=4, x=2 and x=0
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