How do you show that the linearization of #f(x) = (1+x)^k# at x=0 is #L(x) = 1+kx#? Calculus Applications of Derivatives Introduction 1 Answer mason m May 22, 2016 Note that #f(0)=(1+0)^k=1#. Assuming #k# is a constant, we see that #f'(x)=k(1+x)^(k-1)#. Thus #f'(0)=k(1+0)^(k-1)=k*1=k#. Using the point #(0,1)# and slope of #k# we can write the linearization function: #L(x)=k(x+0)+1=1+kx#. Answer link Related questions What is the derivative of the kinetic energy function? What is the derivative of kinetic energy with respect to velocity? What is the derivative of #tanh(x)#? What is the derivative of voltage with respect to time? If a ball is thrown vertically upward from the ground with an initial velocity of 56 feet per... A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first... How do you find the velocity and position vectors if you are given that the acceleration vector... How high will a ball go if it is thrown vertically upward from a height of 6 feet with an... How many seconds will the ball be going upward if a ball is thrown vertically upward from the... How do you find the linearization of #f(x)=x^(3/4)# at x=1? See all questions in Introduction Impact of this question 20375 views around the world You can reuse this answer Creative Commons License