Question

The critical values of x for f(x)=2x^3-15x^2+36 x are?

Answer 1

Answer: 3 and 2

First, find the derivative of the given function

$f(x)=2x^3-15x^2+36x$$f^{\prime}(x)=6x^2-30x+36$

Next, solve for x

$f^{\prime}(x)=6x^2-30x+36$

*Factor out 6 and equate to 0

$6(x^2-5x+6)=0$$6(x-3)(x-2)=0$

*Divide both sides by 6.

$(x-3)(x-2)=0$

*Set x - 3 = 0

$x-3=0$$x=3$

*Set x - 2 = 0

$x-2=0$$x=2$

Next, substitute these x values to the original function.

* x = 3

$x=3$$f(x)=2x^3-15x^2+36x$$f(3)=2(3)^3-15(3)^2+36(3)$$f(3)=2(27)-15(9)^{}+36(3)$$f(3)=54^{}-135+108$$f(3)=27$

* x = 2

$x=2$$f(2)=2(2)^3-15(2)^2+36(2)$$f(2)=2(8)^{}-15(4)^{}+36(2)$$f(2)=16^{}-60^{}+72$$f(2)=28$

We now have a set of critical points

( 3, 27 ) and ( 2, 28 )

The critical values of x would be 3 and 2.