#### Answer

$y=6x-2$

#### Work Step by Step

The question asks us to find the equation of the tangent of the function
$h(t)=t^3+3t$ at the point (1,4)
Given that we are looking for the equation of a tangent, the first reasonable step would be to find the gradient at the points given, which can be found using the derivative function.
Using the power rule:
$h(t)=t^3+3t$
$f′(x)=3t^{3-1}+3=3t^2+3$
$f′(1)=3(1)^2+3=6$
Thus, the gradient of that function at the value x=1 is 6.
Using this information, we can find the intercept value of the tangent line.
The equation of the gradient is given by:
$\frac{y_{1}−y_{0}}{x_{1}−x_{0}}=m$
We can now use the points (1,4) and (0,c) (where c is the y-intercept).
$\frac{4−c}{1-0}=6$
Solving for c gives us c=-2
Thus, the equation of the tangent at (1,4) is $y=6x-2$