Question

y=2x^3-9x^2+12x+5 it has minima at x=2 and maxima at x=1. my question is any other maxima or minima can be found except those point?

Answer 1

Answer on Question #50807 – Math – Differential Calculus | Equations

$y = 2x^3 - 9x^2 + 12x + 5$ it has minima at $x = 2$ and maxima at $x = 1$. My question is any other maxima or minima can be found except those point?

Solution

$$y = 2x^3 - 9x^2 + 12x + 5.$$

Local extrema of differentiable functions can be found by Fermat's theorem, which states that they must occur at critical points. One can distinguish whether a critical point is a local maximum or local minimum by using the first derivative test, second derivative test, or higher-order derivative test, given sufficient differentiability.

According to a Definition of a Critical Number:

if

1) $y(x)$ is not differentiable at $c$, or

2) $y'(x) = 0$,

then $c$ is a critical number of $y(x)$.

1) Function $y(x) = 2x^3 - 9x^2 + 12x + 5$ is differentiable on the entire real line ($(-\infty, +\infty)$).

2) $y'(x) = 6x^2 - 18x + 12$.

Set $y'(x)$ equal to 0:

$$y' = 6x^2 - 18x + 12 = 0 \Rightarrow x^2 - 3x + 2 = 0 \Rightarrow x_1 = 1, x_2 = 2 \text{ are critical numbers}.$$

Using the Second Derivative Test, let's find the relative extrema for $y(x) = 2x^3 - 9x^2 + 12x + 5$.

Using $y'' = 12x - 18 = 6(2x - 3)$ we can apply the Second Derivative Tests:

So, there are no other maxima or minima points except the above indicated ones.

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