We find this solution by:
1)Taking the derivative of #y=3x^2+e^(1-x)#, then
2)Evaluating the derivative at #x=1#, and finally
3)Use the derivative at #x=1# and a point on the function to find the tangent line.
Let us begin by finding the derivative. The rules of taking derivatives tell us that we can split #d/dx(3x^2+e^(1-x))# into #d/dx(3x^2)+d/dx(e^(1-x))#, making it a little easier. In other words, we can do it piece by piece. We start with #3x^2#:
#d/dx(3x^2)=6x-># using the power rule
#d/dx(e^(1-x))=-1*e^(1-x)=-e^(1-x)-># Using the #e^x# rule and the chain rule
Our combined derivative is now #6x-e^(1-x)#. The next step is to evaluate this at #x=1#:
#6(1)-e^((1)-1)=6-e^(0)=6-1=5#
We get 5, which is the slope of the function at #x=1#. Now, to find the equation of the tangent line, we find a point on the function itself, #y=3x^2+e^(1-x)#. Since we are evaluating the tangent line at #x=1#, we will use that as our point:
#y(1)=3(1)^2+e^((1)-1)#
#y(1)=3+1=4#
Our point is #(1,4)#.
Finally, we move on to finding the tangent line. Remember, the tangent line is straight, so its equation will be of the form #y=mx+b#. We have #y# and #x#, we have the slope, but we don't have #b#, the #y#-intercept. In order to find it, we use the information we have now, namely the point #(1,4)# and the slope of 5:
#4=(5)(1)+b#
#4=5+b#
#-1=b#
The #y#-intercept is -1, making our equation #y=5x-1#.
