Practicing Success
Given that $f^{\prime}(x)>g^{\prime}(x)$ for all $x \in R$ and $f(0)=g(0)$, then |
$f(x)>g(x)$ for all $x \in(0, \infty)$ and $f(x)<g(x)$ for all $x \in(-\infty, 0)$ $f(x)<g(x)$ for all $x \in(0, \infty)$ and $f(x)>g(x)$ for all $x \in(-\infty, 0)$ $f(x)>g(x)>$ for all $x \in(-\infty, 0)$ and $f(x)<g(x)$ for all $x \in(0, \infty)$ none of these |
$f(x)>g(x)$ for all $x \in(0, \infty)$ and $f(x)<g(x)$ for all $x \in(-\infty, 0)$ |
Let $h(x)=f(x)-g(x)$ for all $x \in R$. Then, $h'(x)=f'(x)-g'(x)$ for all $x \in R$ $\Rightarrow h'(x)>0$ for all $x \in R$ $\Rightarrow h(x)$ is increasing on R. But, $h(0)=f(0)-g(0)=0$ ∴ $h(x)>h(0)$ for all $x>0$ and, $h(x)<h(0)$ for all $x<0$ $\Rightarrow f(x)-g(x)>0$ for all $x(0, \infty)$ and, $f(x)-g(x)<0$ for all $x \in(-\infty, 0)$ $\Rightarrow f(x)>g(x)$ for all $x \in(0, \infty)$ and, $f(x)<g(x)$ for all $x \in(-\infty, 0)$ |