# pre-Calculus HELP!

• Aug 24th 2006, 08:16 PM
Calculus HELP!
Do not know what to do for the following:

Find and simplify

f(a+h) - f(a)

for each function

f(x) = 4 - x^2

f(x) = x^2 +1
• Aug 24th 2006, 08:37 PM
Soroban

Do you understand functional notation at all?

Given a function \$\displaystyle f(x)\!:\;f(a+h)\$ means "replace each \$\displaystyle x\$ with \$\displaystyle a+h\$."

Quote:

Do not know what to do for the following:

Find and simplify \$\displaystyle f(a+h) - f(a)\$ for each function:

\$\displaystyle (1)\;f(x) \,= \,4 - x^2\$

\$\displaystyle (2)\;f(x)\,=\, x^2 +1\$

We are asked to find; .\$\displaystyle f(a+h) - f(a)\$

Try to "read" what is says:
. . (a) Find \$\displaystyle f(a+h)\$ . . . replace \$\displaystyle x\$ with \$\displaystyle a+h\$.
. . (b) Then find \$\displaystyle f(a)\$ . . . replace \$\displaystyle x\$ with \$\displaystyle a\$.
. . (c) Then subtract them: .\$\displaystyle f(a+h) - f(a)\$

\$\displaystyle (1)\;f(x)\:=\:4 - x^2\$

. . \$\displaystyle (a)\;f(a+h) \;= \; 4 - (a+h)^2\;=\;4 - (a^2 + 2ah + h^2) \;=\$ \$\displaystyle \;4 - a^2 - 2ah - h^2\$

. . \$\displaystyle (b)\;f(a)\;=\;4 - a^2\$

. . \$\displaystyle (c)\;f(a+h) - f(a)\;=\;(4 - a^2 - 2ah - h^2) - (4 - a^2)
\;= \$ \$\displaystyle 4 - a^2 - 2ah - h^2 - 4 + a^2\$