given that f(x)=x^-4x-5, state whether f(x) has a minimum or maximum value and state what the value is?

Answer 1

Hello

Since this is a parabola with positive A value.

Given the form Ax^2 + Bx + C.

The vertex of this is -b/2a = 4/2 = 2.

Since it opens up -- 2 would be the min value.

Hope this helps

Answer 2

I assume you mean f(x)=x^2-4x-5. If so, then
f(x) has a minimum because the x^2 term is positive.
The minimum occurs when x = -b/2a = 4/2=2.
So minimum is 2^2-4*2 - 5 = -9

Answer 3

Let f(x)=x^2-4x-5

for min value take derivative 2x-4=0
2x=4
x=2

or use x=-b/2a

y=4-8-5=-9 is min value of the function

or you can find this value using k=(4ac-b^2)/4a
which is ordinant value of the vertex.

Answer 4

f(x)=x^2-4x-5

has a minimum at (2, -9)

you can find this 2 ways:

take the derivative and set it equal 0

2x-4 = 0

x = 2 y = -9

OR

complete the square and put the equation in the parabola format:

y = a(x-k)^2 +b

y = (x^2 - 4x +4) - 9

y = (x-2)^2 - 9

min at x=2 and y = -9

Answer 5

To find local extrema, take the first derivative of f(x) and set it equal to 0, then solve.

If you don't have the graph of the equation and you need to know if an extremum is a local max or local min, then take the second derivative of f(x) and see if f''(x) is positive or negative at the extremum. If positive, then you have a local min; if negative, then you have a local max.

Answer 6

f(x) = -x2 – 4x + 5