Chapter 8: Q. 7 (page 692)

Let $f (x) = \frac{1}{1 - x}$ and $g (x) = \frac{1}{a x + b}$ . Show that if $b \neq 0$ , thenrole="math" localid="1649659462059" $g (x) = \frac{1}{b} f (- \frac{a}{b} x) .$

Short Answer

Expert verified

$\frac{1}{b} f (- \frac{a}{b} x) = \frac{1}{b} \times \frac{b}{b + a x} = \frac{1}{a x + b} = g (x)$

Hence proved

Step by step solution

Step 1. Given Information.

We have $f (x) = \frac{1}{1 - x}$ and $g (x) = \frac{1}{a x + b}$ .

Show that $g (x) = \frac{1}{b} f (- \frac{a}{b} x)$ .

Step 2. Explanation.

Substitute $x = - \frac{a}{b} x$ in $f (x)$ .

role="math" localid="1649660408537" $f (- \frac{a}{b} x) = \frac{1}{1 - (- \frac{a}{b} x)} = \frac{1}{1 + \frac{a x}{b}} = \frac{1}{\frac{b + a x}{b}} = \frac{b}{b + a x}$

Multiply $\frac{1}{b}$ to $f (- \frac{a}{b} x)$ .

role="math" localid="1649660532439" $\frac{1}{b} f (- \frac{a}{b} x) = \frac{1}{b} \times \frac{b}{b + a x} = \frac{1}{a x + b} = g (x)$

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Let $f (x) = \frac{1}{1 - x}$ and $g (x) = \frac{1}{a x + b}$ . Show that if $b \neq 0$ , thenrole="math" localid="1649659462059" $g (x) = \frac{1}{b} f (- \frac{a}{b} x) .$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Explanation.

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Letf(x)=11-xandgx=1ax+b. Show that ifb≠0, thenrole="math" localid="1649659462059" gx=1bf-abx.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Explanation.

One App. One Place for Learning.

Most popular questions from this chapter

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Let $f (x) = \frac{1}{1 - x}$ and $g (x) = \frac{1}{a x + b}$ . Show that if $b \neq 0$ , thenrole="math" localid="1649659462059" $g (x) = \frac{1}{b} f (- \frac{a}{b} x) .$