Chapter 5: Q. 118 (page 287)

If $f (x) = a^{x}$ , show that $\frac{f (x + h) - f (x)}{h} = a^{x} . \frac{a^{h} - 1}{h} h \neq 0$ .

Short Answer

Expert verified

To prove $\frac{f (x + h) - f (x)}{h} = a^{x} . \frac{a^{h} - 1}{h}$ , first substitute the values of $f (x), f (x + h)$ and solve to get the expression at the RHS.

Step by step solution

Step 1. Given information.

Consider the given question,

$f (x) = a^{x}, \frac{f (x + h) - f (x)}{h} = a^{x} . \frac{a^{h} - 1}{h} h \neq 0$

Substitute $f (x) = a^{x}, f (x + h) = a^{x + h}$ in the given expression,

Take the LHS,

$\frac{f (x + h) - f (x)}{h} = \frac{a^{x + h} - a^{x}}{h}$

Step 2. Use laws of exponents.

Using laws of exponents, $x^{a . b} = x^{a} + x^{b}$ ,

= \frac{a^{x} . a^{h} - a^{x^{}}}{h} = a^{x} . \frac{a^{h} - 1}{h}

Therefore, $L H S = R H S$ .

Hence, proved.

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If $f (x) = a^{x}$ , show that $\frac{f (x + h) - f (x)}{h} = a^{x} . \frac{a^{h} - 1}{h} h \neq 0$ .

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Use laws of exponents.

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Iffx=ax, show thatfx+h-fxh=ax.ah-1hh≠0.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Use laws of exponents.

One App. One Place for Learning.

Most popular questions from this chapter

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If $f (x) = a^{x}$ , show that $\frac{f (x + h) - f (x)}{h} = a^{x} . \frac{a^{h} - 1}{h} h \neq 0$ .