00:01
So for this problem, we are told that the derivative of a function is given by negative 2x minus 2e to the power of x, and we have that f of 0 equals 3.
00:10
In part a, we're told that the function f has a critical point at x equals negative 1.
00:15
We are asked, at this point, does f have a relative maximum, a relative minimum, or neither? and we are asked to justify our answer.
00:22
So there are a few different ways that we can do this.
00:25
One would be to differentiate the function again.
00:27
And in fact considering we're asked about concavity, we should probably take that approach.
00:34
We'd have f double prime of x would be equal to, well, one way we can approach this is product rule, or alternatively we can expand this out.
00:44
So we could consider it as d by dx of negative 2x e to the power of x minus 2 e to the power of x.
00:52
So, derivative of negative 2x e to the x would be negative 2 e to the x minus 2 x e to the x, and then we have another minus 2 e to the x.
01:05
So this is equal to negative 4, e to x minus 2x e to x, or equal to negative 2, eat the x times x plus 2.
01:22
So what we want to do then is see what is the value of our second derivative at x equals negative 1.
01:31
So we would have negative 2 times 1 over e times negative 1 plus 2, so just positive 1.
01:39
That's less than 0.
01:41
So we have that our function is concave downwards.
01:44
If we have that the function as a critical point and it's concave down, then it's going to be a relative maximum.
01:51
Then, for this problem, we are asked, on what intervals, if any, is the graph of f both increasing and concave down? we are asked to explain our reasoning.
02:01
So, we have that if negative 2x minus 2 is greater than 0, the function will be increasing, since each the x is always greater than 0.
02:11
So that means that will be when negative 2x is greater than positive 2, or we can divide both sides by negative 2, but we need to be careful.
02:21
That flips around the inequality...