Question

What is the range of the function f(x)=6x^2+x-1/2x^2+x-6

Answer 1

Answer:

Range: $\:f\left(x\right)\ge \:-\frac{68}{11}$

Interval notation $[-\frac{68}{11},\:\infty \:)$

Step-by-step explanation:

I got a little confused with the x - 1/2x^2 term and I am taking to to be:

$x\:-\frac{1}{2}x^2$  . If this is incorrect please mention the correct term in your comment and I will edit this post. Or you can re-post the question

We have
$f\left(x\right)=6x^2+x\:-\frac{1}{2}x^2+\:x\:-\:6$

Simplify by grouping like terms:
$=-\frac{1}{2}x^2+6x^2+x+x-6$

$= \frac{1}{2}x^2+6x^2+2x-6$

Add similar elements
$-\frac{1}{2}x^2+6x^2$$= -\frac{1x^2}{2}+\frac{12x^2}{2} = \frac{11x^2}{2}$

So the original function expression becomes
$y=\frac{11}{2}x^2+2x-6$

This is the equation of a parabola which in standard form is $ax^2 + bx + c$

Here
$a=\dfrac{11}{2},\:b=2,\:c=-6$

We can compute the vertex of this parabola using the fact that the x, y coordinates of the vertex are given by
$x_v =-\dfrac{b}{2a}$

Plugging in values for b and a give us
$x_v =-\dfrac{b}{2a} = -\dfrac{2}{2\left(\dfrac{11}{2}\right)} =-\dfrac{2}{11}$

Plugging in this value of $x_v$ into the parabola equation will give the value for $y_v$ which is the function value at the vertex

Plug in $x_v = -\frac{2}{11}$ t find the $y_v$ value:
$y_v=\dfrac{11\left(-\dfrac{2}{11}\right)^2}{2}+2\left(-\dfrac{2}{11}\right)-6$

Simplify
$\dfrac{11\left(-\dfrac{2}{11}\right)^2}{2} =\dfrac{2}{11}$

$2\left(-\dfrac{2}{11}\right) = -\dfrac{4}{11}$

Combining the terms we get
$y_v =$ $\dfrac{2}{11}-\dfrac{4}{11}-6$ $= -\dfrac{68}{11}$

$\textsf{Therefore the parabola vertex is at }$ $\left(-\dfrac{2}{11},\:-\dfrac{68}{11}\right)$
For a parabola of the form $ax^2 + bx + c$ with vertex $\left(x_v,\:y_v\right)$

$\circ \;\mathrm{If}\:a < 0\:\mathrm{the\:range\:is}\:f\left(x\right)\le \:y_v$
$\circ \;\mathrm{If}\:a > 0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v$

Here $a=\frac{11}{2}$ so the range is $f\left(x\right)\ge \:-\dfrac{68}{11}$

Answer range of f(x) is
$f\left(x\right)\ge \:-\dfrac{68}{11}$

In interval notation it is $[-\frac{68}{11},\:\infty \:)$

It is much easier if you visualize it in a graph

Hope that helps