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Notely Maths 12
Class 12 Maths
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1.2 Q-2

Question Statement

For the real-valued function ff defined below, find:

  • (a) f−1(x)\boldsymbol{f}^{-1}(x),
  • (b) f−1(−1)f^{-1}(-1),
  • (c) Verify the identities:
    f(f−1(x))=xandf−1(f(x))=xf(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x

The functions given are:

  1. f(x)=−2x+8f(x) = -2x + 8
  2. f(x)=3x3+7f(x) = 3x^3 + 7
  3. f(x)=(−x+9)3f(x) = (-x + 9)^3
  4. f(x)=2x+1x−1f(x) = \frac{2x + 1}{x - 1}, where x>1x > 1

Background and Explanation

To solve these problems, we need to recall the concept of an inverse function. The inverse of a function ff, denoted f−1f^{-1}, is defined such that: f(f−1(x))=xandf−1(f(x))=xf(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x

This means that applying the function ff to its inverse will give back the input, and similarly, applying the inverse function to f(x)f(x) will return the input. In part (a), we will find the inverse functions of the given functions. In part (b), we will substitute x=−1x = -1 into the inverse function. Part (c) will involve verifying the identities for each function.

Solution

(i) f(x)=−2x+8f(x) = -2x + 8

Finding the inverse:

Given f(x)=−2x+8f(x) = -2x + 8, set y=f(x)y = f(x): y=−2x+8y = -2x + 8

Solve for xx: y−8=−2x⇒x=8−y2y - 8 = -2x \quad \Rightarrow \quad x = \frac{8 - y}{2}

Thus, the inverse function is: f−1(x)=8−x2f^{-1}(x) = \frac{8 - x}{2}

Finding f−1(−1)f^{-1}(-1):

Substitute x=−1x = -1 into f−1(x)f^{-1}(x): f−1(−1)=8−(−1)2=92f^{-1}(-1) = \frac{8 - (-1)}{2} = \frac{9}{2}

Verifying f(f−1(x))=xf(f^{-1}(x)) = x:

Substitute f−1(x)=8−x2f^{-1}(x) = \frac{8 - x}{2} into f(x)f(x): f(f−1(x))=f(8−x2)=−2(8−x2)+8=xf(f^{-1}(x)) = f\left(\frac{8 - x}{2}\right) = -2\left(\frac{8 - x}{2}\right) + 8 = x

Thus, f(f−1(x))=xf(f^{-1}(x)) = x holds true.

(ii) f(x)=3x3+7f(x) = 3x^3 + 7

Finding the inverse:

Given f(x)=3x3+7f(x) = 3x^3 + 7, set y=f(x)y = f(x): y=3x3+7y = 3x^3 + 7

Solve for xx: y−73=x3⇒x=(y−73)1/3\frac{y - 7}{3} = x^3 \quad \Rightarrow \quad x = \left(\frac{y - 7}{3}\right)^{1/3}

Thus, the inverse function is: f−1(x)=(x−73)1/3f^{-1}(x) = \left(\frac{x - 7}{3}\right)^{1/3}

Finding f−1(−1)f^{-1}(-1):

Substitute x=−1x = -1 into f−1(x)f^{-1}(x): f−1(−1)=(−1−73)1/3=(−83)1/3f^{-1}(-1) = \left(\frac{-1 - 7}{3}\right)^{1/3} = \left(\frac{-8}{3}\right)^{1/3}

Verifying f(f−1(x))=xf(f^{-1}(x)) = x:

Substitute f−1(x)=(x−73)1/3f^{-1}(x) = \left(\frac{x - 7}{3}\right)^{1/3} into f(x)f(x): f(f−1(x))=f((x−73)1/3)=3((x−73)1/3)3+7=xf(f^{-1}(x)) = f\left(\left(\frac{x - 7}{3}\right)^{1/3}\right) = 3\left(\left(\frac{x - 7}{3}\right)^{1/3}\right)^3 + 7 = x

Thus, f(f−1(x))=xf(f^{-1}(x)) = x holds true.

(iii) f(x)=(−x+9)3f(x) = (-x + 9)^3

Finding the inverse:

Given f(x)=(−x+9)3f(x) = (-x + 9)^3, set y=f(x)y = f(x): y=(−x+9)3y = (-x + 9)^3

Take the cube root of both sides: y1/3=−x+9⇒x=9−y1/3y^{1/3} = -x + 9 \quad \Rightarrow \quad x = 9 - y^{1/3}

Thus, the inverse function is: f−1(x)=9−x1/3f^{-1}(x) = 9 - x^{1/3}

Finding f−1(−1)f^{-1}(-1):

Substitute x=−1x = -1 into f−1(x)f^{-1}(x): f−1(−1)=9−(−1)1/3=9+1=10f^{-1}(-1) = 9 - (-1)^{1/3} = 9 + 1 = 10

Verifying f(f−1(x))=xf(f^{-1}(x)) = x:

Substitute f−1(x)=9−x1/3f^{-1}(x) = 9 - x^{1/3} into f(x)f(x): f(f−1(x))=f(9−x1/3)=(−[9−x1/3]+9)3=xf(f^{-1}(x)) = f(9 - x^{1/3}) = (-[9 - x^{1/3}] + 9)^3 = x

Thus, f(f−1(x))=xf(f^{-1}(x)) = x holds true.

(iv) f(x)=2x+1x−1f(x) = \frac{2x + 1}{x - 1}, where x>1x > 1

Finding the inverse:

Given f(x)=2x+1x−1f(x) = \frac{2x + 1}{x - 1}, set y=f(x)y = f(x): y=2x+1x−1y = \frac{2x + 1}{x - 1}

Multiply both sides by (x−1)(x - 1): y(x−1)=2x+1y(x - 1) = 2x + 1

Expand and solve for xx: yx−y=2x+1⇒(y−2)x=y+1⇒x=y+1y−2yx - y = 2x + 1 \quad \Rightarrow \quad (y - 2)x = y + 1 \quad \Rightarrow \quad x = \frac{y + 1}{y - 2}

Thus, the inverse function is: f−1(x)=x+1x−2f^{-1}(x) = \frac{x + 1}{x - 2}

Finding f−1(−1)f^{-1}(-1):

Substitute x=−1x = -1 into f−1(x)f^{-1}(x): f−1(−1)=−1+1−1−2=0−3=0f^{-1}(-1) = \frac{-1 + 1}{-1 - 2} = \frac{0}{-3} = 0

Verifying f(f−1(x))=xf(f^{-1}(x)) = x:

Substitute f−1(x)=x+1x−2f^{-1}(x) = \frac{x + 1}{x - 2} into f(x)f(x): f(f−1(x))=f(x+1x−2)=2(x+1x−2)+1(x+1x−2)−1=xf(f^{-1}(x)) = f\left(\frac{x + 1}{x - 2}\right) = \frac{2\left(\frac{x + 1}{x - 2}\right) + 1}{\left(\frac{x + 1}{x - 2}\right) - 1} = x

Thus, f(f−1(x))=xf(f^{-1}(x)) = x holds true.

Key Formulas or Methods Used

  • Inverse Function Formula:
    For a function f(x)f(x), the inverse f−1(x)f^{-1}(x) satisfies the property: f(f−1(x))=xandf−1(f(x))=xf(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x

  • Solving for Inverse:
    To find f−1(x)f^{-1}(x), solve the equation y=f(x)y = f(x) for xx in terms of yy.

Summary of Steps

  1. Find the inverse of the function by solving y=f(x)y = f(x) for xx.
  2. Substitute x=−1x = -1 into the inverse function to find f−1(−1)f^{-1}(-1).
  3. Verify the identity f(f−1(x))=xf(f^{-1}(x)) = x by substituting f−1(x)f^{-1}(x) into f(x)f(x).
  4. Verify the identity f−1(f(x))=xf^{-1}(f(x)) = x by substituting f(x)f(x) into f−1(x)f^{-1}(x).