At $x=\frac{5 \pi}{6}, f(x)=2 \sin 3 x+3 \cos 3 x$ is
The maximum slope of curve $y=-x^3+3 x^2+9 x-27$ is
The functin $f(x)=x^x$ has a stationary point at
The maximum value of $\left(\frac{1}{x}\right)^x$ is
The curves $y=4 x^2+2 x-8$ and $y=x^3-x+13$ touch each other at the point ............. .
The curves $y=4 x^2+2 x-8$ and $y=x^3-x+13$ touch each other at the point $(3,34)$. Given, equation of curves are $y=4 x^2+2 x-8$ and $y=x^3-x+13$
$$\begin{array}{ll} \therefore & \frac{d y}{d x}=8 x+2 \\ \text { and } & \frac{d y}{d x}=3 x^2-1 \end{array}$$
So, the slope of both curves should be same
$\therefore \quad 8 x+2=3 x^2-1$
$$\begin{array}{lr} \Rightarrow & 3 x^2-8 x-3=0 \\ \Rightarrow & 3 x^2-9 x+x-3=0 \\ \Rightarrow & 3 x(x-3)+1(x-3)=0 \\ \Rightarrow & (3 x+1)(x-3)=0 \\ \therefore & x=-\frac{1}{3} \text { and } x=3, \end{array}$$
For $x=-\frac{1}{3}$, $$y=4 \cdot\left(-\frac{1}{3}\right)^2+2 \cdot\left(\frac{-1}{3}\right)-8$$
$$\begin{aligned} & =\frac{4}{9}-\frac{2}{3}-8=\frac{4-6-72}{9} \\ & =-\frac{74}{9} \end{aligned}$$
and for $x=3, y=4 \cdot(3)^2+2 \cdot(3)-8$
$$=36+6-8=34$$
So, the required points are $(3,34)$ and $\left(-\frac{1}{3}, \frac{-74}{9}\right)$.