World problem transformation functions

Question

The function is h(x) = 1/10x^2 +10 what is the maximum height reached by the golf ball ?

Answer 1

It's a little ambiguous as written.. that first term could be (1/10)x^2 or 1/(10x^2) or even 1/(10x^2 + 10) It's probably the first though.

if h(x) =(1/10)x^2+10 then

h(infinity) = (1/10)infinity^2 + 10 = infinity^2 as the maximum height

if h(x) = 1/(10x^2)+10 then

h(0) = 1/0 + 10 = infinity, maximum height is infinity when x=0

if h(x) = 1/(10x^2+10) maximum height is reached when 10x^2=-10 or x^2 =-1, x= + or - i

but that's an imaginary solution, with infinity as the maximum height again.

h is maximized when x is 0, since x^2 must be greater than or equal to zero

maximum height = 1/10 + 10 or 10.1, if you ignore the imaginary value for x

so the maximum height is either infinity squared, infinity or 10.1 depending on how you interpret the first term of h(x)

infinity squared is "uncountable"

infinity is countable

in any event, the maximum height is achieved when x=0 or infinity, depending on how you interpet the function as written

It's possible though that the h(x) function was something different than as given, as it doesn't make sense when applied to a golf ball, which starts at ground level.

the usual height problems use the general formula

h(t)=(a/2)t^2 +vt + ho

where a is usually gravity or -32 ft/sec^2 or 9.6 m/sec^2, v is initial velocity and ho is the initial height

h(x)=(1/10)x^2 + 10 leaves out the initial velocity or treats it as v=0, and (1/10) doesn't match with ft/sec^2 or m/sec^2

Maybe it's some formula for another planet