00:01
For each function, we want to find the instantaneous rate of change with respect to x at an arbitrary value of x.
00:06
So we have the two functions, f of x and g of x.
00:09
So first, let's go into the first step here.
00:11
We're going to find the instantaneous rate of change.
00:14
So that would just be the derivative of these functions.
00:17
So if we find f prime of x, which is the derivative of f of x, we can use the power rule here.
00:24
The derivative of x cubed, we would bring that exponent down.
00:28
And that would be 3x.
00:30
And the exponent would decrease by 1 to a second degree.
00:33
And then the derivative of this constant 3 is just 0.
00:36
So that's our derivative.
00:37
F prime of x equals 3x squared.
00:39
That gives us the instantaneous rate of change for any value of x.
00:46
So that's part one for that function.
00:49
Let's go ahead and do part one for the second function, g of x.
00:52
So g prime of x would be the derivative of 6x.
00:56
That's an understood exponent of 1.
00:59
So 1 times 6 is 6.
01:01
And then our degree goes down to 0.
01:04
And x to the 0 is just 1.
01:05
Minus, we do the power rule again here.
01:09
2 comes to the front.
01:10
And we get 2x to the first power.
01:13
So there's our instantaneous rate of changes for both functions.
01:17
Now let's do part two.
01:18
We want to find the slope of the tangent lines for x equals 1.
01:24
That's the same thing as saying the instantaneous rate of change at x equals 1.
01:28
The rate of change gives the slope of the tangent line.
01:32
So f prime at 1 will tell us the slope of the function at 1.
01:37
So that would be 3 times 1 squared.
01:40
And that just gives us 3.
01:43
So that's the rate of change, or the slope of the tangent line, for f of x.
01:48
And now let's do part two for g of x.
01:50
We would just do g prime of 1, which would just be 6 minus 2 times 1.
01:55
That would give us 6 minus 2, which is 4.
01:59
So there's the instantaneous rate of change for g of x.
02:03
Part three, we want to find the equation of the tangent lines to the graphs at x equals 1.
02:11
So we're going to use point -slope form.
02:14
Y minus y1 equals m times x minus x1, where x1, y1 is a point and m is the slope.
02:23
So we need a point on the graph.
02:28
So we're using this when x equals 1.
02:30
So we're going to let x equal 1.
02:33
So what is f of 1? if we plug 1 into this function, that would be 1 cubed plus 3, which would give us 4.
02:40
So that's what we can plug in for our y1, the 4.
02:46
So this would be y minus 4 equals m is just the slope.
02:49
We already found that to be 3.
02:52
And then x minus our x1 is 1.
02:55
So let's solve this.
02:56
That would be y equals 3x minus 3 when we distribute.
03:01
And then we'd add that 4 over.
03:03
And we get y equals 3x plus 1.
03:09
So that gives our equation of the tangent line at x equals 1.
03:16
Let's do the same thing for the g of x.
03:20
We would need to find the points of g of x.
03:24
So in x equals 1, g of 1 would be 6 times 1 minus 1 squared.
03:32
That would just be 6 minus 1, which is 5.
03:36
So that goes to the point 1, 5.
03:39
So that would be y minus 5, and then equals the slope of 4, and then x minus 1.
03:46
That would give us y minus 5 equals 4x minus 4.
03:52
If we add the 5 over, we'll just go ahead and add that to both sides.
03:56
We get y equals 4x and plus 1.
04:02
So there's the equation of the tangent line for g of x at x equals 1.
04:08
Next, we want to find the equation of the secant lines in the interval from negative 1 to 1.
04:14
So to do that, we would need these are the two x values, negative 1 and 1.
04:22
We need to find those associated y values.
04:25
So we have an x value of negative 1.
04:28
We're going to need that y value.
04:30
And then an x value of 1, we're going to need that y value.
04:32
We've already plugged in 1 into f of x, and it gave us 4.
04:36
So we know we have the point 1, 4.
04:38
Now we need to evaluate f at negative 1.
04:41
That would give negative 1 cubed plus 3, which would be negative 1 plus 3, which is 2.
04:48
So that would be the point negative 1, 2.
04:51
So now we need to find the slope of that secant line.
04:55
The slope would be the difference in the y values, 4 minus 2, divided by the difference in the x values, which would be 1 minus negative 1.
05:04
That would give us 2 over 2, which is 1.
05:09
So now we can again use the same method to find the equation of the line.
05:13
We'll use either point we could use here...