How do you find the points where the graph of the function #f(x)= 4x^3 - 30x^2 +48x + 0# has horizontal tangents and what is the equation?

Question

William Abdalla · Accepted Answer

Points are #(4, -32)# & #(1, 22)# & the horizontal tangents are respectively

#y+32=0# & #y-22=0# Given function: #f(x)=4x^3-30x^2+48x# The slope of tangent #dy/dx# at any point to the given curve is given by differentiating #f(x)# w.r.t. #x# as folows #dy/dx=f'(x)# #=d/dx(4x^3-30x^2+48x)# #=12x^2-60x+48# But the horizontal tangent has zero slope i.e. #dy/dx=0# #therefore 12x^2-60x+48=0# #x^2-5x+4=0# #x^2-x-4x+4=0# #x(x-1)-4(x-1)=0# #(x-1)(x-4)=0# #x=4, 1# setting these values of #x# in given function we get corresponding values of y-coordinates as follows #f(4)=4(4)^3-30(4)^2+48(4)=-32# #f(1)=4(1)^3-30(1)^2+48(1)=22# Hence, the coordinates of the points where tangent is horizontal, are #(4, -32)# & #(1, 22)# Now, the equations of horizontal tangents with slope #m=0# at the points #(4, -32)# & #(1, 22)# are given by following formula #y-y_1=m(x-x_1)# #y-y_1=(0)(x-x_1)# #y-y_1=0# Setting #y_1=-32# & #y_1=22# in above equation we get equations of horizontal tangents as follows #y-(-32)=0# #y+32=0# & #y-22=0#

Elijah Abram · Answer

undefined To find the points where the graph of the function f(x) = 4x^3 - 30x^2 + 48x + 0 has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of f(x) by taking the derivative of each term separately:
f'(x) = 12x^2 - 60x + 48

Next, we set the derivative equal to zero and solve for x:
12x^2 - 60x + 48 = 0

We can simplify this equation by dividing all terms by 12:
x^2 - 5x + 4 = 0

Now, we can factor this quadratic equation:
(x - 4)(x - 1) = 0

Setting each factor equal to zero, we find the possible values of x:
x - 4 = 0  -->  x = 4
x - 1 = 0  -->  x = 1

Therefore, the points where the graph of the function has horizontal tangents are (4, f(4)) and (1, f(1)). The equation of the horizontal tangent lines can be found by substituting these x-values into the original function f(x).