Question 3.5.6: Let f be the exponential function and g be the logarithmic f......
Let f be the exponential function and g be the logarithmic function. Then, find:
(i) (f+g)(1) \qquad(ii) (f g)(1)\qquad (iii) (4 f)(1)\qquad (iv) (3 g)(1)
Step-by-Step
Let f: R \rightarrow R: f(x)=e^{x} and g: R^{+} \rightarrow R: g(x)=\log _{e} x .
Then, \operatorname{dom}(f) \cap \operatorname{dom}(g)=R \cap R^{+}=R^{+} .
(i) (f+g): R^{+} \rightarrow R is given by
\begin{aligned}(f+g)(x) & =f(x)+g(x)=\left(e^{x}+\log _{e} x\right) . \\\therefore \quad(f+g)(1) & =\left(e^{1}+\log _{e} 1\right)=(e+0)=e .\end{aligned}(ii) (f g): R^{+} \rightarrow R is given by
\begin{aligned}(f g)(x) & =f(x) \cdot g(x)=e^{x}\left(\log _{e} x\right) . \\\therefore \quad(f g)(1) & =e^{1}\left(\log _{e} 1\right)=(e \times 0)=0 .\end{aligned}(iii) (4 f): R^{+} \rightarrow R is given by
\begin{aligned}(4 f)(x) & =4 \times f(x)=4 e^{x} . \\\therefore \quad(4 f)(1) & =\left(4 \times e^{1}\right)=4 e .\end{aligned}(iv) (3 g): R^{+} \rightarrow R is given by
\begin{aligned}(3 g)(x) & =3 \times g(x)=3 \times\left(\log _{e} x\right) \\\therefore \quad(3 g)(1) & =3 \times g(1)=3 \times\left(\log _{e} 1\right)=(3 \times 0)=0 .\end{aligned}
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