Question
The function \( f:x\rightarrow x^{2}-4x+k\) is defined for the domain \(x\geq p\) ,where k and p are constants.
(i)Express \(f\left ( x \right )\) in the form ,\(\left ( x+a \right )^{2}+b+k\) where a and b are constants.
(ii)State the range of f in terms of k.
(iii)State the smallest value of p for which f is one-one.
(iv)For the value of p found in part (iii),find an expression for \(f^{-1}\left ( x \right )\) and state the domain of \(f^{-1}\) giving your answer in terms of k.
▶️Answer/Explanation
(i) To express \(f(x)\) in the form \((x + a)^2 + b + k\), we expand the square:
\(f(x) = x^2 – 4x + k = (x^2 – 4x + 4) – 4 + k = (x – 2)^2 – 4 + k = (x – 2)^2 + (k – 4)\)
Therefore, \(f(x)\) can be expressed as \((x + a)^2 + b + k\) with \(a = -2\) and \(b = k – 4\).
(ii) The range of \(f\) represents all possible values of \(f(x)\) as \(x\) varies within its domain. Since \(f(x) = (x – 2)^2 + (k – 4)\), we can observe that the square term \((x – 2)^2\) is always non-negative.
Thus, the smallest value \(f(x)\) can attain is when \((x – 2)^2 = 0\), which occurs when \(x = 2\). In this case, \(f(x) = (2 – 2)^2 + (k – 4) = k – 4\). Hence, the smallest value in the range is \(k – 4\).
Therefore, the range of \(f\) is given by \(f(x) \geq k – 4\) for all \(x\) in the domain \(x \geq p\).
(iii) For \(f\) to be one-to-one (injective), each distinct value in the domain must correspond to a distinct value in the range. This means that no two different \(x\) values can produce the same \(f(x)\) value.
To ensure this, we need to set the smallest value of the range \(k – 4\) to be greater than any other value in the range. In other words, \(k – 4\) must be the minimum value of \(f(x)\) for all \(x\) in the domain \(x \geq p\).
Therefore, the smallest value of \(p\) for which \(f\) is one-to-one is \(p = 2\).
(iv) To find the inverse function \(f^{-1}(x)\), we swap the roles of \(x\) and \(f(x)\) in the equation \(f(x) = (x – 2)^2 + (k – 4)\) and solve for \(x\):
\(x = (f^{-1}(x) – 2)^2 + (k – 4)\)
Let’s solve for \(f^{-1}(x)\):
\(x – (k – 4) = (f^{-1}(x) – 2)^2\)
\(\sqrt{x – (k – 4)} = f^{-1}(x) – 2\)
\(f^{-1}(x) = \sqrt{x – (k – 4)} + 2\)
The domain of \(f^{-1}(x)\) is determined by the range of \(f(x)\). From part (ii), we know that \(f(x) \geq k – 4\). Therefore, for \(f^{-1}(x)\) to be defined, we need \(x – (k – 4) \geq k – 4\), which simplifies to \(x \geq 2k – 4\).
Thus, the domain of \(f^{-1}(x)\) is \(x \geq 2k – 4\).
Question
The functions f and g are defined by
\(f(x)=\frac{4}{x}-2\) for\( x> 0\)
\(g(x)=\frac{4}{5x+2}\) for \(x\geq 0\)
(i)Find and simplify an expression for fg(x) and state the range of fg.
(ii)Find an expression for \(g^{-1}(x)\) and find the domain of \(g^{-1}\).
▶️Answer/Explanation
(i) To find the expression for \(f \circ g\) or \(fg(x)\), we substitute the definition of \(g(x)\) into \(f(x)\):
\(fg(x) = f(g(x)) = f\left(\frac{4}{5x+2}\right) = \frac{4}{\frac{4}{5x+2}} – 2 = 5x.\)
Range of f(x) is greater than zero.
So the simplified expression for \(fg(x)\) is \(5x\). The range of \(fg\) is all real numbers since there are no restrictions on \(x\) in the given expressions.
(ii) To find the expression for \(g^{-1}(x)\) and its domain, we follow these steps:
Replace \(g(x)\) with \(y\):
\(y=\frac{4}{5x+2}.\)
\(5x+2 = \frac{4}{y}.\)
\(5x = \frac{4}{y} – 2.\)
\(x = \frac{4-2y}{5y}.\)
Therefore, the expression for \(g^{-1}(x)\) is:
\(g^{-1}(x) = \frac{4-2x}{5x}.\)
The domain of \(g^{-1}\) consists of 5x>0 \cup 4-2x\neq \geq 0 \Rightarrow x\leq 2.
Theefore,domain of \(g^{-1}\) is 0<x\leq 2.
Question
\(f:x\rightarrow \frac{1}{x^{2}-9}\)
\(g:x\rightarrow 2x-3\)
(i)Find and simplify an expression for g(g(x)).
(ii)Find an expression for \(f^{-1}\left ( x \right )\) and state the domain of \( f^{-1}\)
(iii)Solve the equation \(fg(x)=\frac{1}{7}\)
▶️Answer/Explanation
(i) Expression for \(g(g(x))\):
\(g(x) = 2x – 3\)
\(gg(x) = g(2x – 3) = 2(2x – 3) – 3 = 4x – 9.\)
(ii) Expression for \(f^{-1}(x)\) and its domain:
To find the inverse of a function \(f(x)\), we need to swap the roles of \(x\) and \(y\) and solve for \(x\). So, for \(f(x) = \frac{1}{x^2 – 9}\):
Swap \(x\) and \(y\): \(x = \frac{1}{y^2 – 9}\)
Solve for \(y\): \(y^2 – 9 = \frac{1}{x}\)
Take the square root of both sides: \(y = \pm \sqrt{\frac{1}{x} + 9}\)
The domain of \(f^{-1}(x)\) is determined by the original function \(f(x)\).
In this case, the denominator of \(f(x)\) cannot be zero, so \(x^2 – 9 \neq 0\), which means \(x \neq \pm 3\). Additionally, since we’re taking the square root of \(\frac{1}{x} + 9\), we need \(\frac{1}{x} + 9 \geq 0\), which implies \(x > 0\). Therefore, the domain of \(f^{-1}(x)\) is \(x > 0\) and \(x \neq 3\).
(iii)EITHER:
\(\frac{1}{\left ( 2x-3 \right )^{2}-9}=\frac{1}{7}\)
\(\left ( 2x-3 \right )^{2}=16\) or \(4x^{2}-12x-7=0\)
\(x=\frac{7}{2}\) or \(-\frac{1}{2}\)\
\(x=\frac{7}{2}\)
OR:
\(g(x)=f^{-1}\left ( \frac{1}{7} \right )\)
\(g(x)=4\)
\(2x-3=4\)
\(x=\frac{7}{2}\)