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]}

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