4. (4 points) let ( f(x)=2^{x} ). show that \n\n \frac{f(x+h)-f(x)}{h}=2^{x}left(\frac{2^{h}-1}{h}\right)
4. (4 points) let ( f(x)=2^{x} ). show that \n\n \frac{f(x+h)-f(x)}{h}=2^{x}left(\frac{2^{h}-1}{h}\right)
Answer
Explanation:
Step1: Substitute the function into the formula
Given (f(x) = 2^{x}), then (f(x + h)=2^{x + h}). Substitute into (\frac{f(x + h)-f(x)}{h}), we get (\frac{2^{x + h}-2^{x}}{h}).
Step2: Use the exponent rule (a^{m + n}=a^{m}\cdot a^{n})
Since (2^{x + h}=2^{x}\cdot2^{h}), then (\frac{2^{x + h}-2^{x}}{h}=\frac{2^{x}\cdot2^{h}-2^{x}}{h}).
Step3: Factor out the common term (2^{x})
Factor out (2^{x}) from the numerator: (\frac{2^{x}(2^{h}-1)}{h}=2^{x}\left(\frac{2^{h}-1}{h}\right))
Answer:
We have shown that (\frac{f(x + h)-f(x)}{h}=2^{x}\left(\frac{2^{h}-1}{h}\right)) by substituting (f(x)=2^{x}) and (f(x + h)=2^{x + h}), applying the exponent rule (a^{m + n}=a^{m}\cdot a^{n}), and factoring out the common term (2^{x}).