Find a set of values of x for which $x^2-3x+4$ for a decreasing function.
So $x^2-3x+4 \leq 0$ and I am getting $x < \left(\frac {3}{2}\right) \pm {\sqrt {\left(\frac {-7}{4}\right)}}$
Can anyone tell me if its correct ?
I saw the answer and it was $x < 1.5$
Definition. A function is strictly decreasing on an interval $I$ if for each $x_1, x_2 \in I$ such that $x_1 < x_2$, $f(x_1) > f(x_2)$.
Intuitively, what this means is that as the $x$-values increase, the $y$-values decrease. On a graph, it means that the curve falls as you move to the right.
The graph of the function $f(x) = x^2 - 3x + 4$ is the parabola $y = x^2 - 3x + 4$, which opens upwards since the leading coefficient is positive. Therefore, the function decreases to the left of the vertex, reaches its minimum value at the vertex, and increases to the right of the vertex. We can determine the vertex by completing the square.
\begin{align*} f(x) & = x^2 - 3x + 4\\ & = (x^2 - 3x) + 4\\ & = \left(x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} + 4\\ & = \left(x - \frac{3}{2}\right)^2 + \frac{7}{4} \end{align*}
Thus, the function has vertex $(\frac{3}{2}, \frac{7}{4})$, so the function is strictly decreasing if $x < \frac{3}{2}$, as you can see from the graph.