algebra-precalculus Definition and 12 Threads

To prove the upper bound: Let ##c>0##, divide it into ##f(x)## and the coefficients in the final line of the synthetic division tableau are all non-negative. Thus ##f(x)=(x-c)q(x)+r##, where ##r \geq 0## (since the coefficients are given as all non-negative) and is a constant because it's degree...

According to this textbook: https://www.stitz-zeager.com/szprecalculus07042013.pdf On Page ##236## it says: But I can certainly write a combination of powers inside the absolute value: ##|x+1|, |x^2+x+1|##...etc. In fact I can put the definition of a polynomial inside the absolute value...

So assuming I have a graph of a parent function f(x) and I want to graph for example the function f(2x+1). I need to find a way to manipulate the function f(x) to make it look like the function f(2x+1). For the parent function f(x) I have coordinates of (x , y). And for the function f(2x+1) I...

By definition, ##\sqrt{x^2}= \left| x \right|##. For positive ##x##, such as ##4##, it is quite straightforward: ##\sqrt{4^2}=\sqrt{16}=4##. For negative values, I am more confused: ##\sqrt{(-4)^2}=\sqrt{16}=4##. The answer will always be positive, even if you put in a negative value. So why...

##x^2+ax+y^2+by=-c## ##\Leftrightarrow (x+\dfrac{a}{2})^2+(y+\dfrac{b}{2})^2=-c+\dfrac{a^2}{4}+\dfrac{b^2}{4}## The conditions under which the coefficients of this equation makes a circle: ##-c+\dfrac{a^2+b^2}{4}>0## ##\Leftrightarrow 4c < a^2+b^2## Center of circle: ##(-\dfrac{a}{2}...

##x^{\frac{-1}{2}}-2x^{\frac{1}{2}}+x^{\frac{2}{3}}=0## ##\Leftrightarrow x^{\frac{-1}{2}}(1-2x+x^{\frac{7}{6}})=0## Then I'm at a loss as to what to do next because ##x^{\frac{7}{6}}## can't be factored here in a way to get me ##-2x##.

Both sides are in absolute values, i.e. positive, so the solution is straightforward: ##\left| x+3 \right|= \left| 2x+1\right| \Rightarrow x+3=2x+1 \Rightarrow x=2##. But the solution presents another case: ##x+3 = -(2x+1)##. How is this possible if both sides are in absolute values, i.e...

Just by inspection, the answer is obviously ##-2<x<2##. But I tried calculating it step by step and couldn't get the negative portion of the inequality. For ##x>0##, ##x^2<4 \Rightarrow x<2## . Hence ##0<x<2##. For ##x<0##, ##x^2<4 \Rightarrow x>2##. I flipped the inequality because ##x<0##...

##\dfrac{1}{x}<4## For ##x>0##: ##1<4x \Rightarrow x>\dfrac{1}{4}## For ##x<0##: ##1<4x \Rightarrow \dfrac{1}{4}<x \Rightarrow x<-\dfrac{1}{4}## But the problem is ##x<0## works in the original expression instead of just ##x<-\dfrac{1}{4}##, so from calculations alone I missed...

##x^4=\dfrac{81}{16}## ##x=\pm\dfrac{3}{2}##. But I recently realized there are complex solutions as well: ##16x^4-81=0## ##(4x^2)^2-9^2=0## ##(4x^2+9)(4x^2-9)=0## ##x^2=\dfrac{-9}{4}, x^2=\dfrac{9}{4}## ##x=\pm\dfrac{3i}{2}, x=\pm\dfrac{3}{2}## Intuitively when I see ##16x^4=81##, I see...

##5x-6=5x-6## is defined as an identity because it is true for all values of ##x##. My question is I can further simplify and arrive at ##0=0##, in which case no values of ##x## will work because the variable ##x## itself doesn't exist. Isn't this a contradiction? Or did I violate some rule...

##\sqrt{\dfrac{1}{2x^3y^5}}=\dfrac{1}{\sqrt{2\cdot x^2 \cdot x \cdot y^2 \cdot y^2 \cdot y}}=\dfrac{1}{|x|\cdot |y|\cdot |y| \cdot \sqrt{2xy}}=\dfrac{1}{|x|y^2\sqrt{2xy}}## ##\Rightarrow \dfrac{1}{|x|y^2\sqrt{2xy}} \cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{|x|y^2 \cdot...