help plzzz

xshelx1990

hey
reali stuck on this maths past paper q 4m june 2003 paper 6 (thats the calc. paper)..

the expression x^2 - 2mx has a minimum value as x varies

a) find the minimum value of x^2 - 2mx (give your answer in terms of m)

b) state the value of x for which this minimum value occurs (give your answer in terms of m)

thanks!

Reply 1

KAISER_MOLE

x² - 2mx = (x-m)² - m² (expand out and see)

since and real square must give a positive number, the minimum value of the equation occurs when (x-m)² = 0

so, for the questions

a) -m²

b) when x=m

Reply 2

meef cheese

Do you know how to differentiate? If so, differentiate w.r.t x and set it equal to zero.

Or can you complete the square?

If you can't, try to sketch the curve. If you factorise it you get:
x(x-2m)
This shows you that it will equal zero when x=0 or x=2m. So, the minimum TP will be halfway between those two points, at x=m