00:01
So this is a problem that gives us a function, and we're asked to evaluate the function in a variety of different x values or input values.
00:07
So for example, if our function was g of x equals 2x plus 3, if we're given specific x values or input values, we can substitute those for x in our function to determine its output value or its y value.
00:25
So for example, a function notation, if i wanted to evaluate the function g, that means i'm going to use this function that i've been given, i can be given a specific input or x value.
00:36
So let's say that input was zero.
00:39
That means in my function, 2x plus 3, the x value or the input that they would like to evaluate is when the input is 0.
00:47
So i'm going to replace the x value with 0 and leave everything else the same.
00:53
So instead of 2x plus 3, we now have 2 times 0 plus 3, because 0 is the input i was given to evaluate.
01:00
And then if we work out that algebra, we have 2 times 0, which is 0.
01:05
Plus 3, which just equals 3.
01:09
So therefore, g of 0 equals 3.
01:17
Looking at a different example, we had g of negative 4.
01:23
Again, we do the same thing.
01:25
So we're going to replace, we're going to keep the 2 in the function.
01:28
We're going to replace this x value with the input we've been given, which is negative 4.
01:34
Working that out, 2 times negative 4 is negative 8, plus 3 more gives us negative 5.
01:42
So g, the function g, when the input is negative 4, gives us an output of negative 5.
01:51
And we can continue this process with any input that's been given to us.
01:55
So it could be a number, it could be a letter, it could be an expression.
01:59
So continuing through, let's look at g of negative 7.
02:05
We're going to use our expression, 2 times negative 7 plus 3.
02:10
2 times negative 7 is negative 14, plus 3 more is negative 11.
02:16
So g of negative 7 equals negative 11...