00:01
Section 3 .4, problem number 22, so we're presented with a function, 6 minus x squared.
00:11
So we've got f of x is equal to 6 minus x squared.
00:17
We're asked to find the equation of the secant line that passes through x equal negative 1 and x equal 3.
00:26
Okay.
00:27
So to find the slope of the secant line, that's going to be f at x plus.
00:32
Some delta x minus f of x over delta x so here the change in x delta x is going to be three minus negative one which is four so if i look at the secant line that passes through these two points that's going to be so the slope of that secret line is going to be f at so negative one plus delta x which is four minus f at negative 1 over delta x which is 4 so this just is what f of 3 minus f of negative 1 over 4 f of 3 when you evaluate f at 3 that's going to be 6 minus 9 which is negative 3 and then f at negative 1 that's going to be 6 minus 1 which is 5 all of that over 4 so this is negative 8 over 4.
01:38
So that tells me that the slope of the secant line, so the slope of the secant is m equal negative 2.
01:56
So i just need to write what's the equation of a line that passes through this point and, excuse me, the slope of, the slope is negative 2.
02:05
What is the equation of a line that has that slope that passes through either of these points? so you know when x is negative 1, you know that y is equal to 5.
02:17
In that case, when x is 3, negative 3 is y.
02:21
So i got to find equation that passes through negative 1, 5 with a slope of negative 2 or 3, negative 3 with a slope of negative 2.
02:30
So to find that equation, so slope is negative 2, let's go with that first point, negative 1, 5 passes through the point negative 1 5.
02:40
That tells me that y minus 5.
02:43
Is equal to negative 2 x plus 1 so y is equal to minus 2x minus 2 plus 5 y is equal to minus 2x plus 3 this is the equation of the secret line now we're asked to take this same function f of x equals 6 minus x squared and let's find the equation of the tangent line that passes through x equal negative 1.
03:14
So to find the equation of the tangent line, i must find the derivative.
03:19
So i know that f prime of x is going to be the limit as h approaches 0 of f of x plus h minus f of x over h.
03:36
And so i just need to do some algebra here.
03:38
Let's plug in x plus h...