Find \((2f + g – h)(1)\) and \((2f + g – h)(0)\)
Question
If \(f, g, h\) are real functions defined by
\[ f(x)=\sqrt{x+1} \] \[ g(x)=\frac{1}{x} \] \[ h(x)=2x^2-3 \]then find the values of
\[ (2f+g-h)(1) \]and
\[ (2f+g-h)(0) \]Solution
Given
\[ f(x)=\sqrt{x+1} \] \[ g(x)=\frac{1}{x} \] \[ h(x)=2x^2-3 \]We have
\[ (2f+g-h)(x)=2f(x)+g(x)-h(x) \]Find \((2f+g-h)(1)\)
First evaluate each function at \(x=1\).
\[ f(1)=\sqrt{1+1}=\sqrt2 \] \[ g(1)=\frac11=1 \] \[ h(1)=2(1)^2-3 \] \[ =2-3=-1 \]Therefore,
\[ (2f+g-h)(1) \] \[ =2(\sqrt2)+1-(-1) \] \[ =2\sqrt2+2 \]Find \((2f+g-h)(0)\)
At \(x=0\),
\[ g(0)=\frac10 \]which is not defined.
Therefore,
\[ (2f+g-h)(0) \text{ is not defined} \]Final Answer
\[ (2f+g-h)(1)=2\sqrt2+2 \]and
\[ (2f+g-h)(0) \text{ is not defined} \]