Difference between revisions of "OSH/Example/2"
(→Solution) |
|||
| Line 11: | Line 11: | ||
$$ | $$ | ||
| + | ---- | ||
====Solution 1==== | ====Solution 1==== | ||
$$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ | $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ | ||
| Line 19: | Line 20: | ||
=====Comments===== | =====Comments===== | ||
Too terse. | Too terse. | ||
| − | + | ---- | |
====Solution 2==== | ====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 | 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 | ||
| Line 30: | Line 31: | ||
=====Comments===== | =====Comments===== | ||
Too much detail. Repetitive writing. | Too much detail. Repetitive writing. | ||
| − | + | ---- | |
| − | ====Solution | + | ====Solution 3==== |
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, | 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, | ||
$$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ | $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ | ||
Revision as of 23:43, 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 3
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.
