SpecialK Posts: 5, Reputation: 1 New Member #1 Mar 4, 2007, 01:33 PM
Finding x-intercepts, local maximum, local minimum of graph
How do I find the x-intercepts, local maximums, and local minimums of this graph -->
f(x)=1/9(x-3)^2(x+3)^2

using a graphing calculator?
 incognito Posts: 92, Reputation: 24 Junior Member #2 Mar 4, 2007, 01:44 PM
For the x-intercepts, you would set Y to 0 and solve for x.
And for max and min using a TI-83 Plus, when you're looking at your graph hit "2nd"-"Calc"(also the trace key). It is option 3 and 4 in that window for minimum and maximum.
Hope that helps.
 Guest Posts: n/a, Reputation: Guest #3 Apr 2, 2008, 06:38 PM
Finding a composite fuction and iits domain
 tomekdang Posts: 5, Reputation: 2 New Member #4 Jan 22, 2009, 08:52 AM
How do I find the x-intercepts, local maximums, and local minimums of this graph -->
f(x)=1/9(x-3)^2(x+3)^2
We can find x-intercepts like that:
f(x)=0 <=> 1/9(x-3)^2(x+3)^2 = 0 <=> x = 3 or x = -3
one more thing, if you want to find the local maximum/minimum => you must do like that:
f(x) = 1/9 (x-3)(x+3)(x-3)(x+3)
<=> f(x) = 1/9 (x^2 - 9 )^2
<=> f(x) = 1/9 [(x^2)^2 - 2.9.x^2 +9^2]
<=> f(x) = 1/9 (x^4 -18x^2 +81)
find f'(x)
f'(x) = 1/9(4x^3 - 36x)
find x when f'(x) = 0 <=> x = 0 or x = 3 or x = -3
then find f''(x)
f''(x) = 1/9(12x^2 - 36)
determine local maximum and minimum:
f''(0) = 1/9(12.0^2 -36) = -4 <0 => the local maximum of function at x = -4
f''(3) = 1/9(12.3^2 -36) = 8 >0
f''(-3)= 1/9[12.(-3)^2 - 36] = 8 >0
=> this function have 2 local minimum at x = -3 and x = 3
can find y when you calculate x at f(x)
 tinabooe Posts: 2, Reputation: 1 New Member #5 Sep 24, 2009, 07:10 PM

f(x)=0.2x^4+0.3x^3-0.8x^2+5
 morgaine300 Posts: 6,561, Reputation: 276 Uber Member #6 Sep 24, 2009, 11:05 PM

EDIT: Has been re-posted over here:

(That's why this is confusing.)

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