The Absolute maximum of f(x)

Jun 2010
21
0
Consider the function f(x)=5x^2+8x-5. The absolute maximum of f(x) is____

so far i took the derivative of the given and got this 10x+8, from that i got -4/5.
so what to do from this to get the answer for the absolute maximum(Thinking)
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Consider the function f(x)=5x^2+8x-5. The absolute maximum of f(x) is____

so far i took the derivative of the given and got this 10x+8, from that i got -4/5.
so what to do from this to get the answer for the absolute maximum(Thinking)
Since your \(\displaystyle a\) value is positive, your quadratic will have a minimum at the turning point.

So that means that the absolute maximum will be at one of the endpoints. If there are not any endpoints, then the maximum value is \(\displaystyle \infty\).
 
Last edited:
Jun 2010
21
0
how did u get 5 and do i but the original equation in y = ax^2 + bx+c
 
Jul 2007
894
298
New Orleans
if the coefficent infront of x^2 is positive than you have a min and if it neg then you a max. So since 5 is positive you would have a min
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
"how did u get 5 and do i but the original equation in y = ax^2 + bx+c "

YOU said that the function was y= 5x^2+ 8x- 5. Since the leading coefficient is positive this is a parabola that opens upward. It has no "maximum value".

You could also see this by completing the square: y= 5(x^2+ (8/5)x+ 16/25- 16/25)- 5= 5(x+ 4/5)^2- 16/5- 25/5= 5(x+ 4/5)^2- 41/25.

When x= -4/5, the squared term will be 0 and y(-4/5)= -41/25. Since a square is never negative, y is never less than -41/25 so that is the minimum value. There is no bound on how large y can be and so no maximum.