Mixing
Piece-wise function Differentiable?
$begingroup$
Consider the following function $f : Bbb R to Bbb R$:
$$f(x) = begin{cases} sin x & x ge 0 \ x^2 + e^x & x < 0 end{cases}$$
I was wondering if the above piece-wise function was differentiable over the set of Real numbers R, and how to figure it out.
I know that differentiability implies continuity, so maybe I could check if it was continuous as a starter?
calculus
edited Jan 13 at 22:09
zipirovich
11.3k11731
asked Jan 13 at 21:26
Oliver BeckOliver Beck
224
$endgroup$
add a comment |
$begingroup$
Consider the following function $f : Bbb R to Bbb R$:
$$f(x) = begin{cases} sin x & x ge 0 \ x^2 + e^x & x < 0 end{cases}$$
I was wondering if the above piece-wise function was differentiable over the set of Real numbers R, and how to figure it out.
I know that differentiability implies continuity, so maybe I could check if it was continuous as a starter?
calculus
edited Jan 13 at 22:09
zipirovich
11.3k11731
asked Jan 13 at 21:26
Oliver BeckOliver Beck
224
$endgroup$
add a comment |
$begingroup$
Consider the following function $f : Bbb R to Bbb R$:
$$f(x) = begin{cases} sin x & x ge 0 \ x^2 + e^x & x < 0 end{cases}$$
I was wondering if the above piece-wise function was differentiable over the set of Real numbers R, and how to figure it out.
I know that differentiability implies continuity, so maybe I could check if it was continuous as a starter?
calculus
edited Jan 13 at 22:09
zipirovich
11.3k11731
asked Jan 13 at 21:26
Oliver BeckOliver Beck
224
$endgroup$
Consider the following function $f : Bbb R to Bbb R$:
$$f(x) = begin{cases} sin x & x ge 0 \ x^2 + e^x & x < 0 end{cases}$$
I was wondering if the above piece-wise function was differentiable over the set of Real numbers R, and how to figure it out.
I know that differentiability implies continuity, so maybe I could check if it was continuous as a starter?
calculus
calculus
edited Jan 13 at 22:09
zipirovich
11.3k11731
asked Jan 13 at 21:26
Oliver BeckOliver Beck
224
edited Jan 13 at 22:09
zipirovich
11.3k11731
asked Jan 13 at 21:26
Oliver BeckOliver Beck
224
edited Jan 13 at 22:09
zipirovich
11.3k11731
edited Jan 13 at 22:09
zipirovich
11.3k11731
edited Jan 13 at 22:09
zipirovich
11.3k11731
11.3k11731
asked Jan 13 at 21:26
Oliver BeckOliver Beck
224
asked Jan 13 at 21:26
Oliver BeckOliver Beck
224
asked Jan 13 at 21:26
Oliver BeckOliver Beck
224
224
add a comment |
add a comment |
1 Answer
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$begingroup$
That is a good idea. Now check the $x=0$ point. For the right side of the graph, $sin 0 =0$. For the left side, $0^2+e^0=1$. It is not continuous at $x=0$, and is therefore not differentiable at $x=0$.
answered Jan 13 at 21:37
Michael WangMichael Wang
186115
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
That is a good idea. Now check the $x=0$ point. For the right side of the graph, $sin 0 =0$. For the left side, $0^2+e^0=1$. It is not continuous at $x=0$, and is therefore not differentiable at $x=0$.
answered Jan 13 at 21:37
Michael WangMichael Wang
186115
$endgroup$
add a comment |
$begingroup$
That is a good idea. Now check the $x=0$ point. For the right side of the graph, $sin 0 =0$. For the left side, $0^2+e^0=1$. It is not continuous at $x=0$, and is therefore not differentiable at $x=0$.
answered Jan 13 at 21:37
Michael WangMichael Wang
186115
$endgroup$
add a comment |
$begingroup$
That is a good idea. Now check the $x=0$ point. For the right side of the graph, $sin 0 =0$. For the left side, $0^2+e^0=1$. It is not continuous at $x=0$, and is therefore not differentiable at $x=0$.
answered Jan 13 at 21:37
Michael WangMichael Wang
186115
$endgroup$
That is a good idea. Now check the $x=0$ point. For the right side of the graph, $sin 0 =0$. For the left side, $0^2+e^0=1$. It is not continuous at $x=0$, and is therefore not differentiable at $x=0$.
answered Jan 13 at 21:37
Michael WangMichael Wang
186115
answered Jan 13 at 21:37
Michael WangMichael Wang
186115
answered Jan 13 at 21:37
Michael WangMichael Wang
186115
answered Jan 13 at 21:37
Michael WangMichael Wang
186115
186115
add a comment |
add a comment |
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