Porisma

Worked solution

Restrict a quadratic and sketch its inverse

The function f is defined as f:x↦−x2−x+6f:x\mapsto-x^2-x+6, x∈Rx\in\mathbb R.

(i) Explain why f−1f^{-1} does not exist. mark

The function f has an inverse if its domain is restricted to x<kx<k.

(ii) Find the largest value of k and hence find f−1f^{-1} in a similar form. marks

(iii) Sketch, on the same diagram, the graphs of ff, f−1f^{-1} and ff−1ff^{-1}. marks

(iv) Hence, solve the equation f(x)=f−1(x)f(x)=f^{-1}(x), leaving your answer in exact form. marks

[2013/SRJC/JC1MYE/7]

CJC · Review Problems · Question 1

Answer

(i) Not one-to-one. (ii) k=−1/2k=-1/2, f−1(x)=−1/2−25/4−xf^{-1}(x)=-1/2-\sqrt{25/4-x} for x<25/4x<25/4. (iii) The three open endpoints are (−1/2,25/4)(-1/2,25/4), (25/4,−1/2)(25/4,-1/2), (25/4,25/4)(25/4,25/4); see the sketch in the working. (iv) x=−1−7x=-1-\sqrt7.

(i)

The horizontal line y=4 cuts the unrestricted graph twice. Equivalently, f(−2)=f(1)=4f(-2)=f(1)=4, so f is not one-to-one. Hence its inverse function does not exist on all of R\mathbb R. These are the two methods printed in the key.

How did this part go?

(ii)

Complete the square: y=−(x+12)2+254y=-(x+\frac12)^2+\frac{25}{4}. The vertex is at x=−12x=-\frac12, so the largest boundary is k=−12k=-\frac12.

From (x+12)2=254−y(x+\frac12)^2=\frac{25}{4}-y, obtain x=−12±254−yx=-\frac12\pm\sqrt{\frac{25}{4}-y}. Since x<−12x<-\frac12, choose the minus sign.

Thus f−1:x↦−12−254−xf^{-1}:x\mapsto-\frac12-\sqrt{\frac{25}{4}-x}, for x∈Rx\in\mathbb R, x<254x<\frac{25}{4}. The inverse domain is Rf=(−∞,25/4)R_f=(-\infty,25/4).

How did this part go?

(iii)

Sketch the restricted increasing branch of f with open endpoint (−1/2,25/4)(-1/2,25/4). Reflect it in y=x for the inverse, whose open endpoint is (25/4,−1/2)(25/4,-1/2). The composite ff−1(x)=xff^{-1}(x)=x is defined for x<25/4x<25/4, so its line has open endpoint (25/4,25/4)(25/4,25/4). The sketch below retains these three open endpoints.

How did this part go?

(iv)

On this increasing branch, a common point of f and its inverse lies on y=x. Solve

−x2−x+6=x  ⟹  x2+2x−6=0  ⟹  x=−1±7-x^2-x+6=x\implies x^2+2x-6=0\implies x=-1\pm\sqrt7.

The restriction x<−1/2x<-1/2 retains only x=−1−7\boxed{x=-1-\sqrt7}.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Two panels retain both source sketches. First, the unrestricted parabola y=−x²−x+6 has vertex (−1/2,25/4), and y=4 meets it at x=−2 and x=1. Second, the original equal-scale restricted quadratic, reflected inverse and identity composite retain all three open endpoints (−1/2,25/4), (25/4,−1/2), (25/4,25/4), and their common intersection (−1−√7,−1−√7).
Two panels retain both source sketches. First, the unrestricted parabola y=−x²−x+6 has vertex (−1/2,25/4), and y=4 meets it at x=−2 and x=1. Second, the original equal-scale restricted quadratic, reflected inverse and identity composite retain all three open endpoints (−1/2,25/4), (25/4,−1/2), (25/4,25/4), and their common intersection (−1−√7,−1−√7).
Two panels retain both source sketches. First, the unrestricted parabola y=−x²−x+6 has vertex (−1/2,25/4), and y=4 meets it at x=−2 and x=1. Second, the original equal-scale restricted quadratic, reflected inverse and identity composite retain all three open endpoints (−1/2,25/4), (25/4,−1/2), (25/4,25/4), and their common intersection (−1−√7,−1−√7).

Common mistakes

Use the restricted f in the sketch. The open boundary of the original domain also removes the maximum output.

Solution notes

The original inverse/composite SVG is retained with a new unrestricted horizontal-line-test panel. The full two-page source comparison is recorded in evidence/review-cjc-functions-review.md.

Read the hint