Let $f:\mathbb R\rightarrow \mathbb R$ be a twice continuously differentiable function with $f(t)=f(-t)$, and suppose that its graph is given by the following figure.
Let $F:\mathbb R^2\rightarrow \mathbb R$ and $g:\mathbb R\rightarrow \mathbb R^2$ be defined by $F(x,y)=f(\sqrt{x^2+y^2})$ and $g(s)=F(s/\sqrt2,s/\sqrt2).$ (See the following figure.)
(a.) Determine the set of points in $\mathbb R^2$ where the gradient $\nabla F$ is equal to $\langle 0,0 \rangle.$
(b.) Sketch the graphs of $F$ and $g.$
(c.) Sketch the graph of the integral $I(t)=\int_0^t g(s) \, ds.$
(d.) Sketch the graph of the derivative $\frac{d}{ds} g(s).$
I don't see a way to solve this. I know that I can't get the exact function of $f$ only by its graph, at least not by hand.
I tried to compute the gradient of $F,$ but I got it in terms of the partial derivatives of $f(\sqrt{x^2+y^2}).$ I don't think that's what is being requested.
Because of the form of he figure, it looks like one of those problems where you can find the approximate graph of the derivative or the second derivative or the integral, but even with this plausible approach, I don't see how to solve it.
Note: I only wrote every question for reference. I'd just like a hint or something because I feel like I'm missing something.
Thanks in advance.
(a.) Using the Chain Rule, we have that $$F_x = \frac{xf' \bigl(\sqrt{x^2 + y^2} \bigr)}{\sqrt{x^2 + y^2}} \text{ and } F_y = \frac{yf' \bigl(\sqrt{x^2 + y^2} \bigr)}{\sqrt{x^2 + y^2}}.$$ Consequently, we have that $\nabla F = \langle 0, 0 \rangle$ if and only if $F_x = 0$ and $F_y = 0$ if and only if $x f' \bigl(\sqrt{x^2 + y^2} \bigl) = 0$ and $y f' \bigl(\sqrt{x^2 + y^2} \bigr) = 0.$ From here, I believe that you can use the zero product property of the real numbers in conjunction with the graph of $f(t)$ to find the solutions. (Observe that there will be four cases to consider.)
(b.) For sketching $F,$ it might be helpful to focus on $F(x, 0)$ and $F(0, y)$ first. Other than that, there's really not much of a trick; it's kind of a nasty problem. On the other hand, observe that $$g(s) = F \biggl(\frac s {\sqrt 2}, \frac s {\sqrt 2} \biggr) = f \biggl(\sqrt{\biggl(\frac s {\sqrt 2} \biggr)^2 + \biggl(\frac s {\sqrt 2} \biggr)^2} \biggr) = f \bigl(\sqrt{s^2} \bigr) = f(|s|) = f(s)$$ by hypothesis that $f(t) = f(-t),$ so this is quite nice, as you know the graph of $f(s).$
(c.) Observe that the integral $I(t) = \int_0^t g(s) \, ds = \int_0^t f(s) \, ds$ describes the (signed) area between the $s$-axis and the curve $f(s)$ over the closed interval $[0, t]$ for $t \geq 0.$ One can sketch the function $I(t)$ by approximating this area by the graph of $f(s):$ for $0 \leq t \leq 1,$ the graph of $I(t)$ will be the line $y = t;$ for $1 \leq t \leq 2,$ $I(t)$ is increasing because $I'(t) = f(t)$ is positive and concave down since $I''(t) = f'(t) < 0$ (as $f(t)$ is decreasing).
(d.) Considering that $g(s) = f(s),$ we have that $g'(s) = f'(s).$ On the interval $(-1, 1),$ we have that $g(s) = f(s) = 1$ so that $g'(s) = f'(s) = 0.$ Further, when $f(s)$ is decreasing, sketch $f'(s)$ to be negative, e.g., over the interval $(1, 2),$ $f(s)$ is decreasing almost linearly; then, on the interval $(2, 3.5),$ $f(s)$ is decreasing and then increasing quadratically (and concave up), so $f'(s)$ will be increasing linearly, negative for $2 < s < a$ and positive for $a < s < 3.5$ with a root at $s = a.$