Skip to content
Exercise 1.4 · Q3

Q.Graph the functions f(x)=x3f(x)=x^3 and g(x)=x3g(x)=\sqrt[3]{x} on the same coordinate plane. Find f∘gf\circ g and graph it on the plane as well. Explain your results.

Puducherry TnboardTextbookSubjectiveImportance★★★★★est
40% · 42/104 Questions
✓ Free question

Step 1. f(x)=x3f(x)=x^3 and g(x)=x3=x1/3g(x)=\sqrt[3]x=x^{1/3} are inverse functions of each other: cubing then cube-rooting (or vice versa) returns the original number, for every real xx.

Step 2. Compute (f∘g)(x)=f(g(x))=(x1/3)3=x(f\circ g)(x)=f(g(x))=\left(x^{1/3}\right)^3=x, for every real xx (cube root and cube are exact inverses over all of RR, since both are odd bijections R→RR\to R).

Step 3 (Graph). The graph of f∘gf\circ g is therefore just the straight line y=xy=x -- even though ff and gg individually look like an S-curve and its sideways mirror image, composing them cancels all the curvature.

Step 4 (Explanation). This illustrates the general property f∘f−1=If\circ f^{-1}=I (the identity function): composing any bijection with its own inverse, in either order, always collapses back to y=xy=x, regardless of how curved the original functions look.

✓Final answer

(f∘g)(x)=x(f\circ g)(x)=x for all x∈Rx\in R -- the graph is the line y=xy=x, because g=f−1g=f^{-1} and composing a bijection with its inverse always gives the identity function.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.