# The table below shows some values of a linear function f and an exponential function g. Fill in...

## Question:

The table below shows some values of a linear function {eq}f {/eq} and an exponential function {eq}g {/eq}.

Fill in exact values for each of the missing entries.

x 0 1 2 3 4
f(x) 25 375
g(x) 25 375

## Linear and exponential functions

A linear function defines a linear relationship of two variables and is of the form {eq}f(x) = ax + b {/eq}, where {eq}a, b \in R {/eq} are constants. An exponential function defines an exponential relationship between two variables of the function and is of the form {eq}g(x) = cd^x {/eq} where {eq}c,d \in R {/eq} are constants.

#### Determine f(x)

Since f(x) is a linear function, it is of the form {eq}f(x) = ax + b, a,b \in R {/eq} . We need to determine the constants {eq}a \text{ and } b {/eq}.

Plugging in the values from the table row for f(x) yields:$$f(0) = a \cdot 0 + b = 25 \Rightarrow b = 25,\\ f(3) = a \cdot 3 + b = 375,\\ 3a + 25 = 375,\\ a = \dfrac{350}{3}$$

So {eq}f(x) = \dfrac{350}{3}x + 25 {/eq}. Plugging in {eq}x =1,2,4 {/eq} yields:$$f(1) = \dfrac{350}{3} \cdot 1 + 25 = \dfrac{425}{3} \approx 141.67,\\ f(2) = \dfrac{350}{3} \cdot 2 + 25 = \dfrac{775}{3} \approx 258.33, \\ f(4) = \dfrac{350}{3} \cdot 4 + 25 = \dfrac{1475}{3} \approx 491.67$$

#### Determine g(x)

Since g(x) is an exponential function, it is of the form {eq}g(x) = cd^x, c,d \in R {/eq}. We need to determine the constants {eq}c \text{ and } d {/eq}.

Plugging in the values from the table row for g(x) yields:$$g(0) = cd^0 = 25 \Rightarrow c = 25,\\ f(3) = cd^3 = 375,\\ 25d^3 = 375\\ d = \sqrt[3]{15}$$

So {eq}g(x) = 25(\sqrt[3]{15})^x {/eq}. Plugging in {eq}x = 1,2,4 {/eq} yields:$$g(1) = 25(\sqrt[3]{15})^1 \approx 61.66,\\ g(2) = 25(\sqrt[3]{15})^2 \approx 152.06,\\ g(4) = 25(\sqrt[3]{15})^4 \approx 924.83$$

Therefore the exact values for the missing entries are:

x 0 1 2 3 4
f(x) 25 141.67 258.33 375 491.67
g(x) 25 61.66 152.06 375 924.83