find $f(1)$ if relation between $f'(x)$ and $f(x^2)$ is given If $f(x^2) + f'(x) = 6x^2 +7x+2$,
find the value of $f(1)$.
I tried to differentiate $3$ times and but then $f''''(1)$ is not getting eliminated. Can $f(x)$ be linear?
 A: Let $f(1)=a$ and  $f'(1)=b.$ (The key idea: Stay close to $1$.)  Let $g(t)=f(1+t)$ so that $g'(t)=f'(1+t).$ After replacing $x$ by $1+t,$
the right side becomes $$ 15+19t+6t^2 \tag 1$$
Let $$g(t)=a+bt+ct^2. \tag 2$$
It follows that that
$$ g'(t)=b+2ct. \tag 3 $$
Henceforth, we suppress $t^3$ and higher terms. (Imagine that $t$ is less that $10^{-100}.$)
There is an important wrinkle here. I am assuming that the expression in (2) is exact, not just an approximation. Thus, we get the nice expression in (2). Is this legal? Yes, because we're only interested in the values of $g$ and $g'$ at zero.
Also,
$$f(x^2)=f((1+t)^2)=g(2t+t^2)=a+b(2t+t^2)+c(2t+t^2)^2\ \text{so that}$$
$$f(x^2)=a+2bt+(b+4c)t^2 \tag 4$$
We now equate coefficients in the equation $(4)+(3)=(1)$ to obtain
$$a={13\over 3},\  b={32\over 3}, \ c=-{7\over 6}$$
A: As an outsider, I'm unable to add a comment to Michael Hoppe's post.
Nevertheless, let me remark that a Taylor series about $x=1$ would
make a lot more sense. (About zero, you have an unknown function with an
unknown radius of convergence.)
