Question

f) range
11.

a) sketch ( f(x) = sqrt3{x + 1} )
b) domain:
range:
c) algebraically find ( f^{-1}(x) )

d) sketch ( f^{-1}(x) )
e) domain:
f) range

Explanation:

Step1: Analyze parent cube root function

The parent function is $g(x)=\sqrt[3]{x}$, which has domain and range of all real numbers. $f(x)=\sqrt[3]{x+1}$ is a horizontal shift left 1 unit of $g(x)$.

Step2: Find domain of $f(x)$

Cube roots accept all real inputs, so no restrictions on $x+1$.
Domain: All real numbers, $(-\infty, \infty)$

Step3: Find range of $f(x)$

Cube roots output all real numbers, so shifting horizontally does not change the range.
Range: All real numbers, $(-\infty, \infty)$

Step4: Find inverse $f^{-1}(x)$

Set $y=\sqrt[3]{x+1}$, swap $x$ and $y$:
$x=\sqrt[3]{y+1}$
Cube both sides:
$x^3 = y+1$
Solve for $y$:
$y = x^3 - 1$
So $f^{-1}(x)=x^3 - 1$

Step5: Find domain of $f^{-1}(x)$

Polynomials accept all real inputs.
Domain: All real numbers, $(-\infty, \infty)$

Step6: Find range of $f^{-1}(x)$

Cubic polynomials output all real numbers.
Range: All real numbers, $(-\infty, \infty)$

Step7: Sketching notes

• $f(x)=\sqrt[3]{x+1}$: Passes through $(-1,0)$, $(0,1)$, $(-9,-2)$, $(7,2)$; symmetric to its inverse over $y=x$.
• $f^{-1}(x)=x^3 - 1$: Passes through $(0,-1)$, $(1,0)$, $(-2,-9)$, $(2,7)$; symmetric to $f(x)$ over $y=x$.

Answer:

a) (Sketch: Plot points $(-1,0)$, $(0,1)$, $(-9,-2)$, $(7,2)$ and draw a smooth cube root curve shifted left 1 unit; also draw $y=x$ as a dashed line for reference)
b) Domain: $(-\infty, \infty)$
Range: $(-\infty, \infty)$
c) $f^{-1}(x) = x^3 - 1$
d) (Sketch: Plot points $(0,-1)$, $(1,0)$, $(-2,-9)$, $(2,7)$ and draw a smooth cubic curve shifted down 1 unit; symmetric to $f(x)$ over $y=x$)
e) Domain: $(-\infty, \infty)$
f) Range: $(-\infty, \infty)$