we can write

f(x) = ax + b

g(x)= cx + d

as g(x) = c(x - x1)

Then h(x) = f(x)g(x) is a parabola which meet with x-axis at x0 and x1.

(because h(x0)=f(x0)g(x0)=a(x0-x0)g(x0)=0 and likewise, h(x1)=0)

**Case1) a>0 , c>0**

Investigate **the case1**

We can know that it's impossible for parabola to be tangent to both f(x) and g(x).

**Case2) a<0 , c<0**

Investigate **the case2**

In this case it's impossible for parabola to be tangent to both f(x) and g(x) as well.

**Case3) a>0 , c<0**

Since

parabola is symmetric against x=x-coordinate of vertex and

h(x) passes through (x0,0) and (x1,0) and

x-coordinate of vertex of h(x)=(x0+x1)/2 and

h(x) must be tagent to each of f(x) and g(x),

the graph must be the following

we can get the **First Result:**

Let's consider another condition.

We want to get the y-coordinate of the vertex downward and the make the graph as the following

Explore the situation

Note the situation that an intersection of two lines becomes a vertex of their product, parabola.

Let's the vertex (xt, yt). Then h(xt)=f(xt)=g(xt) and h(xt)=f(xt)g(xt)=f(xt)f(xt)=f(xt).

Therfore f(xt)=0 or 1

If f(xt)=g(xt)=0 then

Let's see the graph

If f(xt)=g(xt)=1 then

Let's see the graph

Let's see the graphs simultaneously.

We can realize that the y-value of the
parabola which we want to find **must be below** the product
of the y-values of two lines.

To be that situation, two y-values of
lines **simultaneously must be below "1"and above "0"**.

we can get the **Second Result:**

**RESULT**

Example 1.

Eample 2.

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