Consider the function h(x)= (1/4)x^4-(5/3)x^3+3x^2+4x. Find all values of x where the tangent line to the graph of y= h(x) is parallel to the line y= 4x+3
I found the derivative but now I don't know what to do from here h'(x)= x^3-5x^2+6x+4
Consider the function h(x)= (1/4)x^4-(5/3)x^3+3x^2+4x. Find all values of x where the tangent line to the graph of y= h(x) is parallel to the line y= 4x+3
I found the derivative but now I don't know what to do from here h'(x)= x^3-5x^2+6x+4
Two lines are parallel if and only if they have the same slopes. That is,
$$x^3-5x^2+6x+4=4$$
In other words,
$$x(x^2-5x+6)=0,$$
or
$$x(x-3)(x-2)=0.$$
Now, I believe everything is clear.