How do you find the critical points for #y=x^4-3x^3+5#?

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Answer 1 · 2017-01-11T22:55:15

# (0,5) # and
#(2.25,-3.54) # (2dp)

# y = x^4 - 3x^3 + 5 #

Differentiating wrt #x# gives:

# dy/dx = 4x^3 - 9x^2 #

At a critical point, #dy/dx=0#

# => x^2(4x - 9) = 0#
#:. x=0,9/4#

When;

# x=0 => y=5 #
# x=2.25 => y~~-3.54 #

So there are two critical points:

# (0,5) # and #(2.25,-3.54) # (2dp)

We we can see on the graph:
graph{y = x^4 - 3x^3 + 5 [-2, 5, -5.81, 9.99]}