What is f(x + h) if f(x) = 10-x²?

  • Thread starter Thread starter nick850
  • Start date Start date
AI Thread Summary
To find f(x + h) for the function f(x) = 10 - x², substitute (x + h) into the function, resulting in f(x + h) = 10 - (x + h)². This expands to f(x + h) = 10 - (x² + 2xh + h²). The discussion also highlights that evaluating f at specific points, like f(2) and f(√10), demonstrates how to find function values and identify roots. The key takeaway is the straightforward substitution method for evaluating functions at different inputs. Understanding this process is essential for solving similar problems in algebra.
nick850
Messages
14
Reaction score
0

Homework Statement



if f(x) = 10-x²
f(x + h) = ?


Homework Equations



not applicable

The Attempt at a Solution



no clue
 
Physics news on Phys.org
Simply plug,

(x+h)

Into all the terms that contain an x.
 
thank you!
 
Yep it's as simple as that!

f(x)=10-x^2

f(x+h)=10-(x+h)^2

f(a)=10-a^2

f(2)=10-2^2=6

f(\sqrt{10})=10-(\sqrt{10})^2=0

I think you get the point :smile:

Oh and if you choose some number a such that f(a)=0, this means that a is a root of the equation y=f(x). Such as that last example with the \sqrt{10}.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top