Question

Let f be the exponential function and g be the logarithmic function. Find $(f+g)\left(1\right)$.

• A. $e$
• B. $1$
• C. $0$
• D. $\frac{1}{e}$

Answer 1

The correct option is A $e$

$f:R\to R$ given by $f\left(x\right)=e^x$ and $g:R^{+}\to R$ given by $g\left(x\right)={\log }_{e}x$

Since $Domain\left(f\right)\cap Domain\left(g\right)=R\cap R^{+}=R^{+}$. Therefore,

$f+g:R^{+}\to R$ is given by

$(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)=e^x+{\log }_{e}x$ for all $x\in R^{+}$

Therefore, $(f+g)\left(1\right)=e^1+{\log }_{e}1=e+0=e$