Over which interval is the graph of f(x) = one-halfx2 + 5x + 6 increasing?

[tex]f(x)=\dfrac{1}{2}x^2+5x+6\\\\\text{The coeficient of}\ x^2\ \text{is the positive number. Therefore the parabola is op}\text{en up}.\\\\\text{If a parabola op}\text{en up, then the graph increasing in interval}\ (h,\ \infty),\\\text{and decreasing in interval}\ (-\infty,\ h).\ \text{Where}\ h\ \text{is a first coordinate of a vertex.}\\\\\text{For}\ y=ax^2+bx+c,\ h=\dfrac{-b}{2a}.\\\\\text{We have}\ a=\dfrac{1}{2}\ \text{and}\ b=5.\ \text{Substitute:}\\\\h=\dfrac{-5}{2\left(\frac{1}{2}\right)}=-\cdot\dfrac{5}{1}=-5.[/tex]

joaobezerra joaobezerra · Answer 2 · 2021-11-09T16:14:32+01:00

Using the first derivative test, it is found that the graph is increasing on the interval [tex](-5, \infty)[/tex].

The function is given by:

[tex]f(x) = \frac{x^2}{2} + 5x + 6[/tex]

It's derivative is:

[tex]f^{\prime}(x) = x + 5[/tex]

Then, we test the signal of the derivative, which is the first derivative test.

[tex]x + 5 \geq 0[/tex]

[tex]x \geq -5[/tex]

Thus, since for [tex]x \geq -5[/tex] the first derivative is positive, the function is increasing in the interval [tex](-5, \infty)[/tex].

For [tex]x < -5[/tex], the first derivative is negative, and thus, the function is decreasing.

The graph sketched at the end of this answer corroborates the answer found using the first derivative test.

A similar problem is given at https://brainly.com/question/13539822