Range of quadratic function $y=ax^2+bx+c$ can be easily calculated by transforming it by completing the square method.
$ax^2+bx+c=$\(a{\left( {x + \frac{b}{{2a}}} \right)^2} - \frac{\Delta }{{4a}}\)
We know that
\({\left( {x + \frac{b}{{2a}}} \right)^2} \ge 0\) (where equality occurs when \(x = - \frac{b}{{2a}}\))
When we multiply the left hand side of the equality, then two cases occur:
Case I: $a>0$
\(a{\left( {x + \frac{b}{{2a}}} \right)^2} \ge 0\)
Case II: $a<0$
\(a{\left( {x + \frac{b}{{2a}}} \right)^2} \le 0\) (The "greater than" equality becomes less than equality since we have multiplied the inequality with a negative number.)
Adding \({ - \frac{\Delta }{{4a}}}\) to both sides we get,
\(a{\left( {x + \frac{b}{{2a}}} \right)^2} { - \frac{\Delta }{{4a}}}\ge\)\({ - \frac{\Delta }{{4a}}}\)
or, $y=ax^2+bx+c \ge $ \({ - \frac{\Delta }{{4a}}}\)
We can clearly conclude from here that the minimum value of the expression is \({ - \frac{\Delta }{{4a}}}\) and it occurs when we put \(x = - \frac{b}{{2a}}\) and there is no maximum value.
Therefore the range of the expression is \[\left[ { - \frac{\Delta }{{4a}},\infty } \right)\]
or, $y=ax^2+bx+c \le $ \({ - \frac{\Delta }{{4a}}}\)
We can clearly conclude from here that the maximum value of the expression is \({ - \frac{\Delta }{{4a}}}\) and it occurs when we put \(x = - \frac{b}{{2a}}\) and there is no minimum value.
Therefore the range of the expression is \[\left( { - \infty , - \frac{\Delta }{{4a}}} \right]\]
Example 1: $y=x^2+x+1$
$\Delta = b^2-4ac = 1^2-4.1.1 = -3$
\({ - \frac{\Delta }{{4a}}}={ - \frac{-3 }{{4.1}}}= \frac {3}{4}\)
Since $a=1>0$, Range is \(\left[ {\frac{3}{4},\infty } \right)\)
Example 2: $y=-x^2+x+1$
$\Delta = b^2-4ac = 1^2+4.1.1 = 5$
\({ - \frac{\Delta }{{4a}}}={ - \frac{5 }{{4.-1}}}= \frac {5}{4}\)
Since $a=1>0$, Range is \(\left( { - \infty ,\frac{5}{4}} \right]\).
$ax^2+bx+c=$\(a{\left( {x + \frac{b}{{2a}}} \right)^2} - \frac{\Delta }{{4a}}\)
We know that
\({\left( {x + \frac{b}{{2a}}} \right)^2} \ge 0\) (where equality occurs when \(x = - \frac{b}{{2a}}\))
When we multiply the left hand side of the equality, then two cases occur:
Case I: $a>0$
\(a{\left( {x + \frac{b}{{2a}}} \right)^2} \ge 0\)
Case II: $a<0$
\(a{\left( {x + \frac{b}{{2a}}} \right)^2} \le 0\) (The "greater than" equality becomes less than equality since we have multiplied the inequality with a negative number.)
Adding \({ - \frac{\Delta }{{4a}}}\) to both sides we get,
For the above case I: $a>0$
\(a{\left( {x + \frac{b}{{2a}}} \right)^2} { - \frac{\Delta }{{4a}}}\ge\)\({ - \frac{\Delta }{{4a}}}\)
or, $y=ax^2+bx+c \ge $ \({ - \frac{\Delta }{{4a}}}\)
We can clearly conclude from here that the minimum value of the expression is \({ - \frac{\Delta }{{4a}}}\) and it occurs when we put \(x = - \frac{b}{{2a}}\) and there is no maximum value.
Therefore the range of the expression is \[\left[ { - \frac{\Delta }{{4a}},\infty } \right)\]
For the case II: $a<0$
\(a{\left( {x + \frac{b}{{2a}}} \right)^2} { - \frac{\Delta }{{4a}}}\le\)\({ - \frac{\Delta }{{4a}}}\)or, $y=ax^2+bx+c \le $ \({ - \frac{\Delta }{{4a}}}\)
We can clearly conclude from here that the maximum value of the expression is \({ - \frac{\Delta }{{4a}}}\) and it occurs when we put \(x = - \frac{b}{{2a}}\) and there is no minimum value.
Therefore the range of the expression is \[\left( { - \infty , - \frac{\Delta }{{4a}}} \right]\]
Example 1: $y=x^2+x+1$
$\Delta = b^2-4ac = 1^2-4.1.1 = -3$
\({ - \frac{\Delta }{{4a}}}={ - \frac{-3 }{{4.1}}}= \frac {3}{4}\)
Since $a=1>0$, Range is \(\left[ {\frac{3}{4},\infty } \right)\)
Example 2: $y=-x^2+x+1$
$\Delta = b^2-4ac = 1^2+4.1.1 = 5$
\({ - \frac{\Delta }{{4a}}}={ - \frac{5 }{{4.-1}}}= \frac {5}{4}\)
Since $a=1>0$, Range is \(\left( { - \infty ,\frac{5}{4}} \right]\).
Range of quadratic functions in composition with other functions:
\[af(x)^2+bf(x)+c\]
If $f(x)$ has the range \(\left( { - \infty ,\infty } \right)\) then we can apply the same formulae derived above. In case the range of $f(x)$ is different from \(\left( { - \infty ,\infty } \right)\) we should not use the above formulae of range.
Look at the following examples to get the concept of calculation of range of such functions:
Example 1: Calculate the range of \(2{\sin ^2}x - \sin x + 1\).
Solution: First we convert the above expression to perfect square form using completing the square method:
\[2{\sin ^2}x - \sin x + 1 = 2{\left( {\sin x - \frac{1}{4}} \right)^2} + \frac{7}{8}\]
Now we know that
\[ - 1 \le \sin x \le 1\]
Add to both sides $-\frac{1}{4}$
\[ - \frac{5}{4} \le \left( {\sin x - \frac{1}{4}} \right) \le \frac{3}{4}\]
Now,
\[0 \le {\left( {\sin x - \frac{1}{4}} \right)^2} \le \frac{{25}}{16}\]
[Be careful with squaring]
Multiply both sides with 2,
\[0 \le 2{\left( {\sin x - \frac{1}{4}} \right)^2} \le \frac{{25}}{8}\]
Add $\frac{7}{8}$ to both sides,
\[\frac{7}{8} \le 2{\left( {\sin x - \frac{1}{4}} \right)^2} + \frac{7}{8} \le \frac{{25}}{8} + \frac{7}{8}\]
or,
\[\frac{7}{8} \le 2{\sin ^2}x - \sin x + 1 \le \frac{{32}}{8}\]
Therefore the range of the expression is \(\left[ {\frac{7}{8},4} \right]\)
Example 2: Calculate the range of \(3{\tan ^2}x - \tan x - 1\)
Solution: Since the range of \(\tan x\) is \(\left( { - \infty ,\infty } \right)\), therefore we can simply apply the formula to calculate range:
\(\Delta = {b^2} - 4ac=1+12=13\)
\({ - \frac{\Delta }{{4a}}}={ - \frac{13 }{{4.3}}}= \frac {-13}{12}\)
Since $a=3>0$, Range is \(\left[ {\frac{-13}{12},\infty } \right)\)
Example 3: Calculate the range of \({x^2} - 3|x| - 2\)
Solution: \({x^2} - 3|x| - 2={|x|^2} - 3|x| - 2=\)\({\left( {\left| x \right| - \frac{3}{2}} \right)^2} - \frac{{17}}{4}\)
We know that,
\[|x| \ge 0\]
Adding $-\frac {3}{2}$ to both sides,
\[|x|-\frac {3}{2} \ge -\frac {3}{2}\]
\[{\left( {\left| x \right| - \frac{3}{2}} \right)^2} \ge 0 \] [Again be careful with squaring]
Adding \( - \frac{{17}}{4}\) to both sides we get,
\[{\left( {\left| x \right| - \frac{3}{2}} \right)^2} - \frac{{17}}{4} \ge - \frac{{17}}{4} \]
Therefore the range of the expression is \(\left[ { - \frac{{17}}{4},\infty } \right)\)
Since $a=3>0$, Range is \(\left[ {\frac{-13}{12},\infty } \right)\)
Example 3: Calculate the range of \({x^2} - 3|x| - 2\)
Solution: \({x^2} - 3|x| - 2={|x|^2} - 3|x| - 2=\)\({\left( {\left| x \right| - \frac{3}{2}} \right)^2} - \frac{{17}}{4}\)
We know that,
\[|x| \ge 0\]
Adding $-\frac {3}{2}$ to both sides,
\[|x|-\frac {3}{2} \ge -\frac {3}{2}\]
\[{\left( {\left| x \right| - \frac{3}{2}} \right)^2} \ge 0 \] [Again be careful with squaring]
Adding \( - \frac{{17}}{4}\) to both sides we get,
\[{\left( {\left| x \right| - \frac{3}{2}} \right)^2} - \frac{{17}}{4} \ge - \frac{{17}}{4} \]
Therefore the range of the expression is \(\left[ { - \frac{{17}}{4},\infty } \right)\)
No comments:
Post a Comment