How do you prove a function has neither a local maximum or minimum?

Well, a local max or min would require a critical point---a place where f ' (x) = 0 or is undefined. Try to show that this can't happen. First off, f ' (x) is always well defined since f is a polynomial. f ' (x) = 101 x^(100) + 51 x^(50) + 1. Since 100 and 50 are even numbers, we know that 101 x^(100) and 51 x^(50) are always bigger than or equal to zero. With the added 1, we can say f '(x) ≥ 1 for all real x. That is, f has no critical points! It can't have any local extrema!

A function has a maximum or minimum point when the slope changes direction. So if , say , the slope is rising ( imagine going up the left side of a curve that has a peak on top, and then goes down on the other side.) The slope is positive , then at the top , it is zero and then changes to negative on the way down. So the maximum point on the curve ( or minimum , if the curve was flipped upside down ,), is where the slope is zero,. The slope of a function at a particular point is the first derivative of the function.So all we have to do, is differentiate the function , make it equal to zero , and see what x value , if any , we get. In our case ........ f(x)=x^101+x^51+x+1 ∂y/∂x = 101 x^100 + 51 x^50 +1 you can see in this case , that it does'nt matter which value we choose for x , the derivative will never equal zero , so there is no maximum or minimum

I'm not sure if this would be a legit proof, but here's my explanation: If you graph f(x)=x^101, f(x)=x^51, and f(x)=x individually, they each do not have a local max nor min. Since the sum of those terms just means that you're adding those y values up to get the final curve, as x approaches negative infinity, y approaches negative infinity, and as x approaches positive infinity, y approaches positive infinity. The constant term, 1, just means the graph is shifted up one unit, making no difference to max and mins. If x is to an odd power, as x approaches negative infinity, y approaches negative infinity, and as x approaches positive infinity, y approaches positive infinity. If x is to an even power, as x approaches negative infinity, y approaches positive infinity, and as x approaches positive infinity, y approaches positive infinity. Now, if you ask me that's a pretty good analyzation for a high school pre calc student who hasn't even got to limits!

f(x) = x^101 + x^51 + x + 1 f'(x) = 101x^100 + 51x^50 + 1 Extrema will occur when f'(x) = 0. 0 = 101x^100 + 51x^50 + 1 Let y = x^50. 0 = 101y^2 + 51y + 1 y = (-51 +\- sqrt(2601 - 404))/202 y = (-51 +\- sqrt(2197))/202 y ≈ -0.4845, -0.0204 So x = y^(1/50) = (y^(1/2))^(1/25). As y < 0, y^(1/2) is complex. Then y^(1/2) = a + bi, where a and b are both non-zero reals. Clearly, (a + bi)^(1/25) is some number (c + di) s.t. (c + di)^25 = (a + bi). If d = 0, though, then b must be 0 (because the reals are a field). As this is not the case, d must be non-zero, and so x = y^50 must be non-real. Therefore, the function has no real-valued extrema.

Take the derivative f´(x) = 101x^100 +51x^50+1 =0 call x^50 =z so 101z^2+51z+1=0 The discriminant is z= ((-51+-sqrt (51^2-404)/2 both roots of z are negative but z= x^50 must be positive as 50 is even so the derivative is never zero so the polinom f(x) has no max nor mi n.

You need to prove that it is always increasing. You can do that by examining the derivative. df / dx = 101x^100 + 51x^50 + 1 This is always positive.

A function has a maximum or minimum point when the slope changes direction. So if , say , the slope is rising ( imagine going up the left side of a curve that has a peak on top, and then goes down on the other side.) The slope is positive , then at the top , it is zero and then changes to negative on the way down. So the maximum point on the curve ( or minimum , if the curve was flipped upside down ,), is where the slope is zero,. The slope of a function at a particular point is the first derivative of the function.So all we have to do, is differentiate the function , make it equal to zero , and see what x value , if any , we get. In our case ........ f(x)=x^101+x^51+x+1 ∂y/∂x = 101 x^100 + 51 x^50 +1 you can see in this case , that it does'nt matter which value we choose for x , the derivative will never equal zero , so there is no maximum or minimum

f(x) = x^101 + x^51 + x + 1 f'(x) = 101x^100 + 51x^50 + 1 Extrema will occur when f'(x) = 0. 0 = 101x^100 + 51x^50 + 1 Let y = x^50. 0 = 101y^2 + 51y + 1 y = (-51 +\- sqrt(2601 - 404))/202 y = (-51 +\- sqrt(2197))/202 y ≈ -0.4845, -0.0204 So x = y^(1/50) = (y^(1/2))^(1/25). As y < 0, y^(1/2) is complex. Then y^(1/2) = a + bi, where a and b are both non-zero reals. Clearly, (a + bi)^(1/25) is some number (c + di) s.t. (c + di)^25 = (a + bi). If d = 0, though, then b must be 0 (because the reals are a field). As this is not the case, d must be non-zero, and so x = y^50 must be non-real. Therefore, the function has no real-valued extrema.

Take the derivative f´(x) = 101x^100 +51x^50+1 =0 call x^50 =z so 101z^2+51z+1=0 The discriminant is z= ((-51+-sqrt (51^2-404)/2 both roots of z are negative but z= x^50 must be positive as 50 is even so the derivative is never zero so the polinom f(x) has no max nor mi n.