Find the absolute max/min?

Question

Find the ABSOLUTE MAX ONLY
f(x)=-6x^2+4x-7




Find the ABSOLUTE MAX/MIN of the next two functions?


f(x)=4-5x^2




f(x)=6x/6x+6 on interval [2,4]



please provide steps because I really want to understand how to do it. If you did it on the calculator I need steps too. Thanks! :)

Answer 1

f(x) = -6x² + 4x - 7

f'(x) = -12x + 4 = 0

-12x = -4

x = 1/3



f(1/3) = -6(1/9) + 4(1/3) - 7 = -19/3

Since the graph of f(x) has concave down due to f''(x) < 0, the maximum value of f(x) is -19/3





f(x) = -5x² + 4

f'(x) = -10x = 0

x = 0

f(0) = 4

Again, the graph is f''(x) < 0, so the maximum value of f(x) is 4





f(x) = x / (x + 1)

f(2) = 2/ (2 + 1)

= 2/3



f(4) = 4/(4 + 1)

= 4/5



Since 4/5 > 2/3, the maximimum value between [2, 4] is 4/5 at x = 4



Hope this helps :D

Answer 2

f(x)=-6x^2+4x-7

a=-6 so you know the parabola is a 'mountain'.

The function reaches an absolute and also relative maximum when the derivative is 0:

f'(x) = 4-12x = 0

x = 1/3

And f(1/3) = -19/3



So the function reaches an absolute maximum -19/3 at x=1/3.

______

For the next function you use the same method to find the maximum and you will find that the function reaches an absolute maximum 4 at x = 0.

There's no absolute minimum.

_______

f(x) = 6x/(6x+6) =(simplify) x/(x+1)

f'(x) = 1/(x+1)²

The derivative is never 0 so there are no relative maxima/minima.

This means the absolute maximum and minimum on the interval [2,4] are the boundary points f(2) and f(4).

f(4) = 4/5 = max

f(2) = 2/3 = min

Answer 3

a. Find f ' (x) (gradient function) by differentiating so we get dy/dx =-12x + 4



and we want to find where it's a maximum, as it's a negative x^2 graph it looks like an "n" so we know the gradient at the maximum is 0



so -12x + 4 = o



-12x = -4



x = 1/3



And y = f(1/3) I can't be bothered to work it out.



I can't be bothered to work the others out. For the second one, the minimum will be -infinity and find the max by doing what I did here.



For the last you will still need to differentiate by using the quotient rule and then if you draw this graph you should see where it's going to be a max between 2 and 4...



Good luck :P always best to keep trying :P

Answer 4

in spite of if it particularly is a quadratic function, the min or max (given by utilising the sign x^2) is calculated by spinoff. F(x)=ax^2+bx+c a-provides the min or max the spinoff is 2ax+b the element of min or max is x = - b/2a