Mathematics
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OpenStudy (lena772):
Evaluate (if possible) the function at the given
value(s) of the independent variable. Simplify the results.
f(x)=x+3/x+1
f(x)-f(1)/x-1
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OpenStudy (lena772):
OpenStudy (lena772):
@ganeshie8
ganeshie8 (ganeshie8):
that red circled one ?
OpenStudy (lena772):
Yes :)
ganeshie8 (ganeshie8):
you're given,
\(\large f(x) = x^3-x\)
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ganeshie8 (ganeshie8):
and asked to evaluate,
\(\large \frac{f(x)-f(1)}{x-1}\)
ganeshie8 (ganeshie8):
start by finding \(f(1)\)
ganeshie8 (ganeshie8):
then plug that value in the expression you wanto evaluate
OpenStudy (lena772):
f(1)=1^3-1
f(1)=0?
ganeshie8 (ganeshie8):
Excellent !
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OpenStudy (lena772):
:)
ganeshie8 (ganeshie8):
plug that value in the given expression
OpenStudy (lena772):
f(x)-0/x-1
f(x)/x-1
ganeshie8 (ganeshie8):
plug f(x) value also, and simplify
ganeshie8 (ganeshie8):
\(\large \frac{f(x)}{x-1}\)
\(\large \frac{x^3-x}{x-1}\)
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ganeshie8 (ganeshie8):
simplify
OpenStudy (lena772):
I don't know how to simplify that
ganeshie8 (ganeshie8):
factor top and see if u can cancel out something
ganeshie8 (ganeshie8):
\(\large \frac{f(x)}{x-1}\)
\(\large \frac{x^3-x}{x-1}\)
\(\large \frac{x(x^2-1)}{x-1}\)
\(\large \frac{x(x+1)(x-1)}{x-1}\)
OpenStudy (lena772):
x(x+1)?
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OpenStudy (lena772):
which would be x^2+x ?
ganeshie8 (ganeshie8):
\(\large \color{red}{\checkmark}\)
OpenStudy (lena772):
so all i have to write for my answer is x^2+x?
ganeshie8 (ganeshie8):
yup, that will do
OpenStudy (lena772):
can you help me with another?