00:01
In this question, we are asked to find an equation of the tangent line, which is common to the graphs of the functions f and g.
00:08
And let's say the tangent line touches the graph of f at the point x ,0, y not.
00:26
And it's tangent, it touches the graph of g at the point x1 and y1.
00:40
Now, the slope of the tangent line to f is f prime of x0.
00:48
Right? on the other hand, the slope of the line is g prime of x1, right? g prime of x1 is also the slope of the tangent line.
01:24
Now, since it's the same line, the slopes must be equal.
01:32
F prime of x not must be equal to g prime of x1, because it's the same line.
01:51
So its slope must be constant.
01:53
Now f prime of x not equal, f prime of x not equal, f prime of x, equals to 2x.
02:02
G prime of x equals to 2 times x minus 5.
02:11
Therefore, f prime of x0 equals to 2 times x0 and g prime of x1 equals to 2 times x1 minus 5.
02:26
After canceling 2, we are going to get that x0 equals to x1 minus 5.
02:34
Therefore, the relation between the points at which the tangent line touches the graphs of the functions is given by x0 equals to x1 minus 5.
02:48
Let's remember that.
02:50
Now let's write down an equation of the tangent line to f at the point x0 .y not.
03:00
So the equation of the tangent line to f at the point x0y0 is y minus y0 equals to f prime of x0 times x minus x0.
03:29
Well, we just found that f prime of x is 2x, therefore f prime of x not equals to 2x0.
03:39
F prime of x0 is 2x0 and y0 is simply f of x not.
03:49
And recall that f of x equals to x squared.
03:52
So f of x not equals to x not squared.
03:56
And we can rewrite the equation of the tangent line as y minus x not squared equals to 2 .8.
04:02
X0 multiplied by x minus x not what we are going to do next is replace x not by x1 minus 5 if we replace x0 by x1 minus 5 we are going to get y minus x1 minus 5 squared equals to 2 times x1 minus 5 multiplied by x minus x1 minus 5.
04:50
So what we did is just replaced x0 everywhere in the equation of the tangent line by x1 minus 5.
04:59
Now recall that this tangent line touches the graphs of both functions.
05:06
And it touches the graph of g at the point x1y1, right? which means that the point x1 y1 belongs to the line.
05:13
The point x1 y1 is on the line and this means that we can plug in y equals to y1 in the equation of the tangent line and x equals to x1 so we can plug in x equals x1 and y equals to y1 because the point x1 and y1 belongs to the line right and after doing that what we are going to get is y1 minus x1 minus 5 squared equals to 2 times x1 minus 5 multiplied by x1 minus x1 plus 5.
06:14
Then we can cancel x1 here.
06:18
And now y1 equals to g of x1.
06:27
And if we remember that g of x equals to x minus 5 squared plus 20, we are going to get that g of x equals...