Question

question 22 · 1 point
let $f(x) = 5(4)^{x - 1}$. evaluate $f(3)$ without using a calculator.
provide your answer below:
$f(3) = \square$
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question 23 · 1 point
estimate $\sqrt{255}$ to two decimal places.
provide your answer below:
$\sqrt{255} \approx \square$

Explanation:

Question 22

Step1: Substitute \( x = 3 \) into \( f(x) \)

Substitute \( x = 3 \) into the function \( f(x)=5(4)^{x - 1} \), we get \( f(3)=5(4)^{3 - 1} \).

Step2: Simplify the exponent

Simplify the exponent \( 3-1 = 2 \), so the expression becomes \( f(3)=5(4)^{2} \).

Step3: Calculate \( 4^{2} \)

Calculate \( 4^{2}=16 \).

Step4: Multiply by 5

Multiply 5 by 16, \( 5\times16 = 80 \).

Step1: Find the perfect squares around 255

We know that \( 15^{2}=225 \) and \( 16^{2}=256 \). Since \( 255 \) is very close to \( 256 \) (which is \( 16^{2} \)), we can start with that.

Step2: Use linear approximation (or observe the difference)

The difference between \( 256 \) and \( 255 \) is \( 256 - 255=1 \). The derivative of \( y = \sqrt{x} \) at \( x = 256 \) is \( y^\prime=\frac{1}{2\sqrt{x}} \), at \( x = 256 \), \( y^\prime=\frac{1}{2\times16}=\frac{1}{32}\approx0.03125 \). Using linear approximation \( \sqrt{255}\approx\sqrt{256}-0.03125\times1 = 16 - 0.03125 = 15.96875 \). Rounding to two decimal places, we get \( 15.97 \). (Alternatively, since \( 255=256 - 1 \), and \( \sqrt{256 - 1}\approx16-\frac{1}{32}=15.96875\approx15.97 \))

Answer:

\( f(3)=\boxed{80} \)

Question 23