What is the equation of the line tangent to # f(x)=3x^2 + e^(1-x)# at # x=1#?

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Answer 1 · 2015-12-20T06:11:27

#y = 5x -1#

Step 1: Differentiate the function
#f'(x) = d/dx(3x^2 ) + d/dx(e^(1-x))#

#= 6x -e^(1-x)#

Step 2: Find the slope of tangent line at #x= 1#
# m= f'(1) = 6(1) -e^(1-1) = 5#

Step 3: Find #y# coordinate of the function when #x= 1#

#f(1) = 3(1)^2 + e^(1-1) = 3 + 1 = 4#

So, original point of the graph is #(1,4)#

To find equation of the line we can us point slope formula

#y- y_1 = m(x-x_1)# , when #x_1# & #y_1# is the point on original equation

#m= 5# #" " " " " " " (1, 4) #

# y - 4 = 5(x-1)#
#=> y = 5x -5+4#

#=> y = 5x -1#