The composite function works like this:
So, if we substitute back z = 3x, the final output is 1/(3x+2). We know that the denominator of a fraction can't be zero, so we must impose
top problems:
1-) x²-2x-8
<h2>(x-4)(x+2)</h2>---------------------
2-) 4s²-4s+1
<h2>(2s-1)²</h2>---------------------
3-) 6x²-3-7x
<h2>(2x-3)(3x+1)</h2>---------------------
4-) 4u²+8u
<h2>4u(u+2)</h2>---------------------
5-) t²-15
<h2>2t</h2>---------------------
6-) 3x²-x-4
<h2>(3x-4)(x+1)</h2>bottom problems:
1-)
y=x²+
---------------------
2-)
---------------------
3-) 0, √5
<h2>y=x²-x√5</h2>---------------------
4-) 1,1
<h2>y= x²-2x+1</h2>---------------------
5-)
---------------------
6-) 4,1
<h2>y=x²-5x+4</h2>