If $g(x)=3^x+2x^{\frac{1}{2}}$. Then prove that the equation $g'(x)=21$ has at least one real roots
What i try:: given $g(x)=3^{x}+2x^{\frac{1}{2}}$
Then $\displaystyle g'(x)=3^x\ln(3)+2\cdot \frac{1}{2}x^{-\frac{1}{2}}=3^x\ln(3)+\frac{1}{\sqrt{x}}$
Now i did not understand How do i find number of solution of $g'(x)=21$
Help me please , thanks
Best Answer
$g'$ is continuous at $(0,+\infty)$.
$$g'(1)=3\ln(3)+1<21$$
$$\lim_{x\to +\infty}g'(x)=+\infty$$
thus there exist $ b>1$ such that $g'(b)>21$.
$21\in (g'(1),g'(b)) $ so, By EVT, $$\exists c\in (1,b) \; : g'(c)=21$$
Remark:
$g' $ is a derivative, we do not need its continuity.