Use linear approximation, i.e. the tangent line, to approximate 15.3^2 as follows:
Let f(x)=x^2 and find the equation of the tangent line to f(x) at x=15. Using this, find your approximation for 15.3^2

Question

Use linear approximation, i.e. the tangent line, to approximate 15.3^2 as follows: 
Let f(x)=x^2 and find the equation of the tangent line to f(x) at x=15. Using this, find your approximation for 15.3^2

anonymous · Answer

The linear approximation of $f(x)$ at $x=15$ is given by the line
$$g(x)=f(15)+f'(15)(x-15),$$
where $f(x)=x^2$.

So, the approximate value of $15.3^2$ is given by $g(15.3)\approx f(15.3)$.

anonymous · Answer

start with f(x)

anonymous · Answer

a linear approximation means, for a small increment in x, estimate the change in y to a degree of "1"
i.e., tangent line at that point
so, our point of function is at x="15"

anonymous · Answer

$$f(x)=x^2\implies f(15)=225$$
now, for a small change in 'x', (dx), find change in f (f')
$$dy=2xdx\implies \Delta y=2x\Delta x$$
and near x=15,
$$\Delta y=2\times15\times\Delta x=13\Delta x$$
now, our number is "15.3", which differs by 0.3 (=dx) from 15(=x)
$$\Delta y=30(0.3)=9\
f(x+\Delta x)=y+\Delta y\
f(15.3)=f(15)+\Delta y=225+9=234
$$

anonymous · Answer

thank u!

anonymous · Answer

welcome