For what values of #m# is #F(x) = (m^2+m+1)x# increasing or decreasing?

Question

Abigail Allen · Accepted Answer

undefined To determine if the function $ F(x) = (m^2 + m + 1)x $ is increasing or decreasing, we need to examine the sign of its derivative with respect to $ x $, $ F'(x) $. If $ F'(x) > 0 $, the function is increasing. If $ F'(x) < 0 $, the function is decreasing. To find $ F'(x) $, we need to differentiate $ F(x) $ with respect to $ x $: $$ F'(x) = (m^2 + m + 1) $$ The sign of $ F'(x) $ is determined by the sign of $ m^2 + m + 1 $. If $ m^2 + m + 1 > 0 $, $ F(x) $ is increasing. If $ m^2 + m + 1 < 0 $, $ F(x) $ is decreasing. We can use the discriminant $ \Delta = b^2 - 4ac $ to determine the sign of the quadratic expression $ m^2 + m + 1 $. If $ \Delta < 0 $, the quadratic expression has no real roots, and since the coefficient of $ x^2 $ is positive, the quadratic expression is always positive. Therefore, $ F(x) $ is always increasing. If $ \Delta > 0 $, the quadratic expression has two real roots, and the sign of $ m^2 + m + 1 $ depends on the value of $ m $. If $ \Delta = 0 $, the quadratic expression has one real root, and again, the sign of $ m^2 + m + 1 $ depends on the value of $ m $. Therefore, for all real values of $ m $, the function $ F(x) = (m^2 + m + 1)x $ is always increasing.

Genesis Allen · Answer

#F(x)# is strictly increasing for any #m in RR#...

Note that: #m^2+m+1 = m^2+2(1/2)m+1/4+3/4# #color(white)(m^2+m+1) = (m+1/2)^2+3/4 >= 3/4# for any #m in RR# Hence #F(x)# is the equation of a line with positive slope, thus strictly increasing.

Eva Abramson · Answer

undefined To determine for what values of $m$ the function $F(x) = (m^2 + m + 1)x$ is increasing or decreasing, we need to analyze the sign of the coefficient of $x$, which is $m^2 + m + 1$.

If $m^2 + m + 1 > 0$, then $F(x)$ will be increasing for all $x$.
If $m^2 + m + 1 < 0$, then $F(x)$ will be decreasing for all $x$.

The discriminant of the quadratic equation $m^2 + m + 1$ can help us determine its sign. The discriminant $b^2 - 4ac$ for the quadratic equation $ax^2 + bx + c = 0$ is used to determine the nature of the roots. If the discriminant is positive, the roots are real and distinct. If it's zero, the roots are real and equal. If it's negative, the roots are complex.

For $m^2 + m + 1$, the discriminant is $1^2 - 4(1)(1) = 1 - 4 = -3$. Since the discriminant is negative, $m^2 + m + 1$ is always positive, regardless of the value of $m$.

Therefore, the function $F(x) = (m^2 + m + 1)x$ is always increasing for all values of $m$.