Question

Solving Equations and Inequalities Graphically Graphs ofthe functions f and g are given.

(a) Which is larger, $f\left(0\right)$or $g\left(0\right)$)?

(b) Which is larger, $f\left(-3\right)$or $g\left(-3\right)$?

(c) For which values of x is $f\left(x\right)=g\left(x\right)$?

(d) Find the values of x for which $f\left(x\right)=g\left(x\right)$.

(e) Find the values of x for which $f\left(x\right)>g\left(x\right)$.

Short answer

1. $f\left(0\right)>g\left(0\right)$
   
2. $f\left(-3\right)<g\left(-3\right)$
   
3. At $x=-2$and $-4$when$f\left(x\right)=g\left(x\right)$
4. At$x=2,3,-2,-3$$f\left(x\right)\le g\left(x\right)$
5. At$x=0,1,-1$ the value of$f\left(x\right)>g\left(x\right)$

Step-by-step solution

Part a. Step 1. Determining the abscissa and ordinate.

Here, the-axis represent the abscissa and –axis y-axis represent the ordinate.

Part a. Step 2. Concept of comparison of two function at a point.

Let the function$f$ and $g$the point is $a$.

So, if $f\left(a\right)=b$and$g\left(a\right)=c$

Where $b>c$.

Then $f\left(a\right)>g\left(a\right)$otherwise$f\left(a\right)<g\left(a\right)$

Part a. Step 3. Comparison of f0 and g0.

At$x=0,f\left(0\right)=2$

And At$x=0,0g\left(0\right)<1$

So,$f\left(0\right)>g\left(0\right)$

$f\left(0\right)$is larger than$g\left(0\right)$

Part b. Step 1. Determining the abscissa and ordinate.

Here, the-axis represent the abscissa and –axis y-axis represent the ordinate.

Part b. Step 2. Concept of comparison of two function at a point.

Let the function$f$ and $g$the point is $a$.

So, if $f\left(a\right)=b$and$g\left(a\right)=c$

Where $b>c$.

Then $f\left(a\right)>g\left(a\right)$otherwise$f\left(a\right)<g\left(a\right)$

Part b. Step 3. Comparison of f-3 and g-3.

At $x=-3,f\left(-3\right)=-2$

And At $x=-3,g\left(-3\right)>2$

So, $f\left(-3\right)<g\left(-3\right)$

$g\left(-3\right)$is greater than$f\left(-3\right)$

Part c. Step 1. Determining the abscissa and ordinate.

Here, the-axis represent the abscissa and –axis y-axis represent the ordinate.

Part c. Step 2. Concept of finding x where fx=gx.

When the function$f$ and $g$intersect each other at any point $a$then at $f\left(a\right)=g\left(a\right)$.

Part c. Step 3. Determining the value of x where fx=gx.

At$x=-2$$f$ and $g$intersect at $y=1$.

$\therefore f\left(-2\right)=g\left(-2\right)=1$.

And At $x=2$$f$and $g$intersect at$y=2$

$\therefore f\left(-2\right)=g\left(-2\right)=2$.

Hence, at $x=-2$and $2$$f\left(x\right)=g\left(x\right)$

Part d. Step 1. Determining the abscissa and ordinate.

Here, the-axis represent the abscissa and –axis y-axis represent the ordinate.

Part d. Step 2. Concept of finding x when fx≤gx.

When the function$f$ and $g$intersect each other at any point $a$then at $f\left(a\right)=g\left(a\right)$.

Part d. Step 3. Determining the value x for which when fx≤gx.

At$x=0,f\left(0\right)>g\left(0\right)$

At$x=1,f\left(1\right)>g\left(1\right)$

At$x=2,f\left(2\right)=g\left(2\right)$

At$x=3,f\left(3\right)<g\left(3\right)$

At$x=-1,f\left(0\right)>g\left(-1\right)$

At$x=-2,f\left(-2\right)=g\left(-2\right)$

At$x=-3,f\left(-3\right)<g\left(-3\right)$

And at$x=2,3,-2,-3,f\left(x\right)\le g\left(x\right)$

Part e. Step 1. Determining the abscissa and ordinate.

Here, the-axis represent the abscissa and –axis y-axis represent the ordinate.

Part e. Step 2. Concept of finding x when fx>gx.

Let at$f\left(a\right)=b$ and$g\left(a\right)=c$ such that$b>c$ then $f\left(a\right)>g\left(a\right)$.

Part e. Step 3. Determining the value x for which when fx>gx.

At$x=0,f\left(0\right)>g\left(0\right)$

At$x=1,f\left(1\right)>g\left(1\right)$

At$x=2,f\left(2\right)=g\left(2\right)$

At$x=3,f\left(3\right)<g\left(3\right)$

At$x=-1,f\left(0\right)>g\left(-1\right)$

At$x=-2,f\left(-2\right)=g\left(-2\right)$

At$x=-3,f\left(-3\right)<g\left(-3\right)$

So, at$x=0,1,-1,f\left(x\right)>g\left(x\right)$