The equation of the tangent at (x, y) to the given curve y = f(x) is
$$Y - y = {{dy} \over {dx}}(X - x)$$
Y-intercept $$ = y - x{{dy} \over {dx}}$$
According to the question
$${x^3} = y - x{{dy} \over {dx}}$$
$$ \Rightarrow {{dy} \over {dx}} - {y \over x} = - {x^2}$$
which is linear in x.
$$IF = {e^{\int {{{ - 1} \over x}dx} }} = {1 \over x}$$
$$\therefore$$ Required solution is
$$y\,.\,{1 \over x} = \int { - {x^2}\,.\,{1 \over x}dx} $$
$$ \Rightarrow {y \over x} = {{ - {x^2}} \over 2} + c$$
$$ \Rightarrow y = {{ - {x^3}} \over 2} + cx$$
At x = 1, y = 1,
$$1 = {{ - 1} \over 2} + c$$
$$ \Rightarrow c = {3 \over 2}$$
Now, $$f( - 3) = {{27} \over 2} + {3 \over 2}( - 3)$$
$$ = {{27 - 9} \over 2} = 9$$
If $$g$$ is a function defined on $$R$$ with values in the interval $$\left( {0,\infty } \right)$$ such that
$$$f\left( x \right) = ln\,\left( {g\left( x \right)} \right),\,\,for\,\,all\,\,x \in R$$$
then the number of points in $$R$$ at which $$g$$ has a local maximum is $$1$$.