Question

The graph shows the function f(x). On a coordinate plane, a cube root function goes through (8, 2), has an inflection point at (0, negative 1), and goes through (8, negative 3). Which equation represents f(x)? f(x) = Negative RootIndex 3 StartRoot x EndRoot f(x) = Negative RootIndex 3 StartRoot x minus 1 EndRoot f(x) = Negative RootIndex 3 StartRoot negative x EndRoot minus 1 f(x) = Negative RootIndex 3 StartRoot negative x EndRoot

Answer 1

Answer:

f(x) = -∛x   - 1

Step-by-step explanation:

x     y = f(x)

8   -3

0    -1

=> f(x) = -∛x   - 1

=> f(8)   = -∛8  - 1

=> f(8) = -2  - 1

=> f(8) = -3

f(0) =  -∛0  - 1

=> f(0) = -0 - 1

=> f(0) = -1

hence correct equation is

f(x) = -∛x   - 1

Answer 2

Given:

F(x)= $\sqrt[3]{x}$

point (8, 2)

To prove:

F(x)= ?

Solution:

Given function:

$i)\ \ \ f(x) =-\sqrt[3]{x} \\\\ ii) \ \ \ f(x)= -\sqrt[3]{x-1}\\\\ iii) \ \ \ f(x)= -\sqrt[3]{x}-1\\\\ iv) \ \ \ f(x)= -\sqrt[3]{-x}$

put value 8 in the x and check it will give value that is equal to 2.

$i)\ \\ f(x)= -\sqrt[3]{8}\\\\f(x) = -\sqrt[3]{2^3}\\\\f(x) = -2 \ that's \ why \ it \ is \ wrong.$

$ii)\ \\ f(x)= -\sqrt[3]{8-1}\\\\f(x) = -\sqrt[3]{7}\\\\f(x) = \ its \ value \ lie \ in \ between -2 to -3\ that's \ why \ it \ is \ wrong.$

$iii) \\ f(x)= -\sqrt[3]{8}-1\\\\f(x) = -\sqrt[3]{2^3}-1\\\\f(x) =-2-1\\\\ f(x) =-3 \ that's \ why \ it \ is \ wrong$

$iv) \\ f(x)= -\sqrt[3]{-8}\\\\f(x) = -\sqrt[3]{-2^3}\\\\f(x) =-(-2)\\\\ f(x) =2$

Option iv) that is $f(x)= -\sqrt[3]{-x}$ is correct.