Need help with local max/min?

Question

I have this math problem due and I've done it twice but I'm still not 100% it's right.
The original equation is f(x)= x^3 - (11/2)x^2 + 10x - 19
for the derivative I got 3x^2 - 11x + 10
so my critical points are x=2 and x= (5/3)

It asks when f(x) is increasing and I think its negative infinity to postive infinity, but that could be wrong haha

And lastly it asks where the local max and/or min is and I'm lost

So if anyone could help that would be amazing!

MechEng2030 · Accepted Answer

f '(x) = 3x^2 - 11x + 10 = 0

3x^2 - 6x - 5x + 10 = 0

3x(x - 2) - 5(x - 2) = (3x - 5)(x - 2) = 0

x = 5/3, 2

f '(x) is increasing where f '(x) > 0

(3x - 5)(x - 2) > 0

It's increasing for x > 2 and x < 5/3.

Via second derivative test:

f "(x) = 6x - 11

f "(5/3) < 0 (Therefore x = 5/3 is a local max).

f "(2) > 0 (Therefore x = 2 is a local min).

V.G.Panneerselvam · Answer

f(x)=x^3-(11/2)x^2+10x-19

f '(x)=3x^2-11x+10=(3x-5)(x-2)

f "(x)=6x-11

condition for local maxima or minima is for x=c, f '(x)=0 and if f "(x)<0 then f(x)is maximum

at x=c. if f "(x)>0, then f(x) is minimum at x=c.

here at x=5/3 and x=2, f '(x)=0

at x=5/3, f "(x)=6(5/3)-11=10-11 = -1 <0, thus at x=5/3 f(x) is local maximum

at x=2 f (x)=6.2-11=12-11 = 1>0 then at x=2, f(x)is local minimum.

Trending News

Need help with local max/min?

2 Answers