1. ## derivatives

if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).

2. Originally Posted by asnxbbyx113
if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
Point-Slope Form?

f(1) = 2 gives the point
f'(1) = 5 gives the slope.

y - 2 = 5(x-1)

Substitute x = 1.2

3. Originally Posted by asnxbbyx113
if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
Or you can think of it as a truncated Taylor series.
$f(1.2) \approx f(1) + \frac{1}{1!}f^{\prime}(1) \cdot 0.2$

So
$f(1.2) \approx 2 + 5 \cdot 0.2$
etc.

-Dan

4. Originally Posted by topsquark
Or you can think of it as a truncated Taylor series.
$f(1.2) \approx f(1) + \frac{1}{1!}f^{\prime}(1) \cdot 0.2$

So
$f(1.2) \approx 2 + 5 \cdot 0.2$
etc.

-Dan
That is not the best explanation to give this poster. Because he will not hear the term "Taylor Series" for another year.