"If $\displaystyle f(x) = x^2 - 3x$, determine the values of $\displaystyle x$ for which $\displaystyle f(x - 3) \leq 0$"
This is for only two marks, but I'm not quite sure how to go about doing this... Any tips would be great!
"If $\displaystyle f(x) = x^2 - 3x$, determine the values of $\displaystyle x$ for which $\displaystyle f(x - 3) \leq 0$"
This is for only two marks, but I'm not quite sure how to go about doing this... Any tips would be great!
$\displaystyle f(x-3) = (x-3)^2 - 3(x-3) = (x-3)^2 - 3x +9$
Can you find the values of X for which the above expression is $\displaystyle \leq 0$
hint: Write it as a quadratic then find the solutions that give exactly 0. Then plot the graph and the range of values is clear
Spoiler:
Or if you didn't want to waste time graphing, completing the square solves that quadratic inequality beautifully...
$\displaystyle \displaystyle \begin{align*}(x - 3)^2 - 3x + 9 &\leq 0 \\ x^2 - 6x + 9 - 3x + 9 &\leq 0 \\ x^2 - 9x + 18 &\leq 0 \\ x^2 - 9x + \left(-\frac{9}{2}\right)^2 - \left(-\frac{9}{2}\right)^2 + 18 &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 - \frac{81}{4} + \frac{72}{4} &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 - \frac{9}{4} &\leq 0 \\ \left(x - \frac{9}{2}\right)^2 &\leq \frac{9}{4} \\ \left|x - \frac{9}{2}\right| &\leq \frac{3}{2} \\ -\frac{3}{2} \leq x - \frac{9}{2} &\leq \frac{3}{2} \\ 3 \leq x &\leq 6 \end{align*} $