Difference between revisions of "OSH/Example/2"
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| − | ====Solution==== | + | ====Solution 1==== |
| − | As the component parts of $f$ are continuous on their own, the only potential discontinuity is at $x=1$. Calculating the limit from the left, | + | $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ |
| + | $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ | ||
| + | $$1+a=3$$ | ||
| + | $$a=2$$ | ||
| + | |||
| + | =====Comments===== | ||
| + | Too terse. | ||
| + | |||
| + | ====Solution 2==== | ||
| + | The component parts of $f$ are polynomials and we saw in class (and in the course notes - page 458, line 12) that polynomials are continuous. This means that away from $x=1$, this function is continuous. The only point that whose continuity is in question is the point $x=1$. To check continuity at this point, we calculate the limit of $f(x)$ as $x$ approaches 1 from the left and get | ||
| + | $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ | ||
| + | This last step is justified because as I already mentioned, the function is just a polynomial on either side of $x=1$ and so continuous. That means we can evaluate the left sided limit simply by evaluating the function at the limiting value. | ||
| + | Next we calculate the limit of $f(x)$ as $x$ approaches 1 from the right, | ||
| + | $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ | ||
| + | Here again, the last step is justified by the fact that the function is a polynomial on either side of $x=1$ and hence continuous. That means we can evaluate the left sided limit simply by evaluating the function at the limiting value. For $f$ to be continuous, we require that $1+a=3$. Solving this for $a$ we find that <span style="border: solid 1px grey;padding: 3px 3px 3px 3px;">$a=2.$</span> | ||
| + | |||
| + | =====Comments===== | ||
| + | Too much detail. Repetitive writing. | ||
| + | |||
| + | ====Solution 2==== | ||
| + | As the component parts of $f$ are polynomials and hence continuous on their own, the only potential discontinuity is at $x=1$. Calculating the limit from the left, | ||
$$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ | $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ | ||
And from the right, | And from the right, | ||
$$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ | $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ | ||
For $f$ to be continuous, we require that $1+a=3$ or that <span style="border: solid 1px grey;padding: 3px 3px 3px 3px;">$a=2.$</span> | For $f$ to be continuous, we require that $1+a=3$ or that <span style="border: solid 1px grey;padding: 3px 3px 3px 3px;">$a=2.$</span> | ||
| + | |||
| + | =====Comments===== | ||
| + | This is the style you should emulate. | ||
Revision as of 23:42, 11 September 2013
Example problem statement
For what value of $a$ is the function $f(x)$ continuous at all points, where $f(x)$ is the following piecewise function $$ f(x) = \left\{ \begin{array}{ll} x^2+ax & \quad x < 1 \\ 4-x & \quad x \geq 1 \end{array} \right. $$
Solution 1
$$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ $$1+a=3$$ $$a=2$$
Comments
Too terse.
Solution 2
The component parts of $f$ are polynomials and we saw in class (and in the course notes - page 458, line 12) that polynomials are continuous. This means that away from $x=1$, this function is continuous. The only point that whose continuity is in question is the point $x=1$. To check continuity at this point, we calculate the limit of $f(x)$ as $x$ approaches 1 from the left and get $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ This last step is justified because as I already mentioned, the function is just a polynomial on either side of $x=1$ and so continuous. That means we can evaluate the left sided limit simply by evaluating the function at the limiting value. Next we calculate the limit of $f(x)$ as $x$ approaches 1 from the right, $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ Here again, the last step is justified by the fact that the function is a polynomial on either side of $x=1$ and hence continuous. That means we can evaluate the left sided limit simply by evaluating the function at the limiting value. For $f$ to be continuous, we require that $1+a=3$. Solving this for $a$ we find that $a=2.$
Comments
Too much detail. Repetitive writing.
Solution 2
As the component parts of $f$ are polynomials and hence continuous on their own, the only potential discontinuity is at $x=1$. Calculating the limit from the left, $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ And from the right, $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ For $f$ to be continuous, we require that $1+a=3$ or that $a=2.$
Comments
This is the style you should emulate.
