Also: Explain why the function is discontinuous at the given number a.
f(x) = (1)/(x + 1)
a = −1
What is the domain of x=b^y?
2015-12-07 9:29 pm
回答 (3)
2015-12-07 9:32 pm
b^y is always greater than 0
f(-1) is undefined.
f(-1) is undefined.
2015-12-07 9:43 pm
(a) Consider that y is a function of x.
x=b^y=>
yln(b)=ln(x)=>
y=ln(x)/ln(b)
=>
dom(y)={x | x is real & x>0}
(b) f(x)=1/(x+1)
limit f(x)=infinity
x->-1
Infinity is not a finite nor definite value, so
f(x) is discontinuous at x=-1, By definition,
if f(x) is continuous at x=a, then
limit f(x)=f(a)
x->a
where f(a) exists & is a finite & definite value.
x=b^y=>
yln(b)=ln(x)=>
y=ln(x)/ln(b)
=>
dom(y)={x | x is real & x>0}
(b) f(x)=1/(x+1)
limit f(x)=infinity
x->-1
Infinity is not a finite nor definite value, so
f(x) is discontinuous at x=-1, By definition,
if f(x) is continuous at x=a, then
limit f(x)=f(a)
x->a
where f(a) exists & is a finite & definite value.
2015-12-07 9:33 pm
Don't main
收錄日期: 2021-05-01 15:27:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151207132900AAgej1z