Determining local extrema and absolute extrema. Please help by 12am! Thanks!?

Question

Here's my two problems:
g(x) = x^2 + 1

Options:
a) local min: 1; absolute min: 1
b) local min: -1; absolute min: -1
c) no local extrema; absolute min: 1
d) local max: 1; absolute max: 1

ALSO

g(x) = - |x+6|

Options:
a) local min: 0; absolute min: 0
b) local max: -6; absolute max: -6
c) local max: 0; absolute max: 0
d) local max: 6; absolute max: 6

Is this a calculator problem or something else? Grrr teacher didn't explain this one in class so I'm a bit confused on the whole concept.

Answer 1

a) depends upon what you mean be a local versus absolute min. An absolute min is also a local min.

So g(x)=x^2+1
so g(x) has a stationary point where dg/dx=0. Which is the same as g'(x)=0 depending upon your notation.

This is the same as 2x=0, which is at x=0, where g(x)=1.

g''(x)=2 so g(0) is a maximum.

g(x)=-|x+6| is a tricky question because it is not differentiable at g(-6) which is also the maximum. Instead you could show it is strictly concave around the root of x=-6.

To show this, you need to show that for any two points u,v where u
or -t|u|-6t+(t-1)|v|+(t-1)6 <=-|tu+(1-t)v|-6

-t|u|+(t-1)|v| -6<=-|tu+(1-t)v|-6

multiplying both sides by -1
t|u|+(1-t)|v|>=|tu + (1-t)v|

You can then show by cases this must be true where u,v<-6, u<-6,v=-6,u=-6,v>-6, u,v>-6.

Likewise, you can show that as the limit of g(x) as x approaches infinity approaches - infinity and the limit of g(x) as x approaches negative infinity approaches - infinity. Likewise the limit of g(x) as x approaches -6 from the left g(x) approaches 0, likewise the same can be shown from the right,

Therefore -6 is a maximum.