Which of these is a critical point of the function
f(x) = 1/2 x^4 - 3x^3 + 1/2 x^2 + 12x + 7
in the interval [-1.5, 2]?

Question

Which of these is a critical point of the function 
f(x) = 1/2 x^4 - 3x^3 + 1/2 x^2 + 12x + 7
in the interval [-1.5, 2]?

jango_in_dtown · Answer

First find the derivative of f and then set it to zero. solve and get the values of x and you have the critical points

jango_in_dtown · Answer

f'(x)=??

welshfella · Answer

differentiate term by term using the rule 

if f(x) = ax^n ,  f'(x) =  anx^(n-1)

3mar · Answer

Critical points of f(x) are when f'(x)=0 or when f'(x) is undefined

3mar · Answer

@love_to_love_you Do you follow?

love_to_love_you · Answer

finally, now I can read y'all's responses. Sorry, my computer wasn't working

love_to_love_you · Answer

-1.5
-1
1
1.25


Those are the choices, I graphed the function, is there any way I can determine what the critical point is through the graph?

welshfella · Answer

Yes the critical point will be a turning point on the graph

solomonzelman · Answer

Suppose (for instance) that you want to find the critical points of
$\color{black}{f(x)=(1/3)x^3-(5/2)x^2+6x-4}$ over the interval $\color{black}{x \in (e,10]}$.

THEORY:
Critical points of any function $\color{black}{f(x)}$ (in general) are possible in 3 cases:
 $\color{black}{\small [1]}$  If $\color{black}{f'(a)=0}$, then $\color{black}{f'(x)}$ has a critical point (CP) at $\color{black}{x=a}$.
 $\color{black}{\small [2]}$  If $\color{black}{f'(a)}$ is undefined (i.e. DNE) and $\color{black}{f(a)}$ exists, then $\color{black}{f'(x)}$ has a CP at $\color{black}{x=a}$.
 $\color{black}{\small [3]}$  Interval's closed boundary is a CP.

SOLUTION:
$\color{black}{f(x)=(1/3)x^3-(5/2)x^2+6x-4}$
$\color{black}{f'(x)=x^2-5x+6}$
$\color{black}{0=x^2-5x+6}$
$\color{black}{0=(x-2)(x-3)}$
So, $\color{black}{x=2}$  and  $\color{black}{x=3}$.
(Note that our interval starts from $\color{black}{x=e}$ so the only critical point for $\color{black}{f(x)}$ (thus far) is $\color{black}{x=3}$.

Note also that $\color{black}{f(x)}$, and by definition (as well as by common sense, and all world's logic) polynomials are always differentiable. (In other words, $\color{black}{f'(x)}$ is defined (as well as differentiable) everywhere). So, case [2] excluded.

Now, we had two interval boundaries, and since $\color{black}{x=10}$ is the only included boundary (only closed boundary) your second critical point is $\color{black}{x=10}$.

So, the answer is: $\color{black}{x=3,~10}$.

solomonzelman · Answer

I meant to say that ...
note also that f(x) is a polynomial.

solomonzelman · Answer

If you have any questions (conceptual, practical or any), please ask and we will address them.

solomonzelman · Answer

If you haven't learned calculus, then the only way is graphing.

Closed interval boundaries, sharp angles (within the interval, if any is given), and points of horizontal tangencies, are all the critical points.

love_to_love_you · Answer

On my graph, (-1, -1) is a turning point. Would -1 be my answer?

solomonzelman · Answer

there are two other solutions.

solomonzelman · Answer

Yes, (-1,-1) is correct.

solomonzelman · Answer

oh my bad, I didn't look at the interval.

solomonzelman · Answer

this is the only solution ... good job.

love_to_love_you · Answer

Thank you

love_to_love_you · Answer

Thanks everyone for helping

solomonzelman · Answer

$y \omega$