Question: Find the number of solutions of equation $f(x)=2^x-x^2+x+\cos x=0$
Given answer is $1$
My Attempt
I have checked the derivative of it. $$f'(x) = 2^x\ln2-2x+1-\sin x\\f''(x)=2^x(\ln2)^2-2-\cos x$$ but $f'(x)$ is both +ve and -ve so I could not make any conclusion. I had projected this graph in desmos and...
The function is not monotonic. Now I did this. $$\begin{align} x+ \cos x &= x^2-2^x \\ g(x) &=h(x) \end{align}$$ $g(x)= x+\cos x \implies g'(x)=1-\sin x\gt 0\; ; \forall x \in \mathbb R\implies g(x)$ is increasing
$h(x)= x^2-2^x \implies h'(x)=2x-2^x\ln2 \implies h''(x)=2-2^x(\ln2)^2$; $\;h''(x)=0 \implies $ $x= 1-2\log_2 \ln2\\ x \approx 2.0575$
for this value of $x \; ,\;$ $h'(x) \gt 0$ and $\begin{cases}x\to-\infty & h'(x)\to-\infty\\ x\to\infty & h'(x)\to-\infty \end{cases}$ thus $h'(x)=0$ for two values of $x$ (but seems to me impossible to find which two values without graphical calculator). So finally I could not make any comment about $h(x)$ from here also. How to solve this question without any graphing calculator?