We have discussed functions and relations on our page Functions and Relations, now we will discuss value of function for given x.
To calculate value of given function at x=a, we plug in a for x in the expression.
And then simplify the expression to get the value.
For example, if we have $f(x)=2x^2+3x+5$
For example, if we have $f(x)=2x^2+3x+5$
Let's calculate value of function f(x) at x=2.
We have $f(2)=2\cdot 2^2+3\cdot2+5=8+6+5=19$
Similarly, we can plug any value for x, even we can substitute any algebraic expression for x in the expression.
If we have $f(x)=2x^2+5x-7$
Let's calculate $x=2+h$, we get
If we have $f(x)=2x^2+5x-7$
Let's calculate $x=2+h$, we get
$f(x)=2x^2+5x-7$
$f(2+h)=2(2+h)^2+5(2+h)-7$
$= 2(4+4h+h^2)+10+5h-7$
$=8+8h+2h^2+10+5h-7$
Adding like terms, we get
$f(2+h)=2h^2+13h+11$
$f(2+h)=2h^2+13h+11$
and $f(2)=(2\cdot 2^2+5\cdot 2-7)=11$
And we can calculate, $f(2+h)-f(2)$
$f(2+h)-f(2)\\=(2h^2+13h+11)-11\\=2h^2+13h+11-11\\=2h^2+13h$
Now we can divide it by h, and we get
$\dfrac{f(2+h)-f(2)}{h}=\dfrac{2h^2+13h}{h}=2h+13$
$\dfrac{f(2+h)-f(2)}{h}=2h+13$
This expression is called Difference Quotient.
Difference Quotient is helpful in finding and understanding a derivative at a point.
It's used in limit and derivative topics extensively.
It's used in limit and derivative topics extensively.
Try these problems, find the difference quotient for each problem.
1. $f(x)=5x^2+2x-9$ at x=3
1. $f(x)=5x^2+2x-9$ at x=3
2. $f(x)=3x^2-2x+9$ at x=2
3. $f(x)=x^2-2x+3$ at x=1
Sometimes, we say that difference quotient is the slope of tangent to the curve of f(x) at x=a where a be any real number.
Slope of tangent to any curve f(x) at x=a is $slope=\dfrac{f(a+h)-f(a)}{h}$.
Then, we can obtain equation of tangent using point slope form as $y=m(x-a)+f(a)$ and here point is (a, f(a)) and slope m is the difference quotient and $m=\dfrac{f(a+h)-f(a)}{h}$
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