Problem with this question
$$f(x)=3\sin x\cos \frac x 2 \text{ around } a=1$$
I have been breaking up $f(x)$ into $g(x) = 3\sin x$ and $h(x)= \cos \frac x 2$, and then multiplying them back ($g(x)\cdot h(x)$) this was suggested to my by the lecturer. but for the life of me I cant get the answer.
Answer is: $y = 0.817x+1.398$
any help would be greatly appreciated.
You apparently only need the 1st-degree Taylor polynomial; note that the 1st-degree Taylor polynomial for a function $f(x)$ at $x=a$ is the same as the tangent line to the graph of $f$ at $a$, $y=f(a)+f'(a)\cdot(x-a)$.
Given your $g$ and $h$, you can write 1st-degree Taylor polynomials for each of them, then take the product, as you said. For $g$, $$y=g(1)+g'(1)\cdot(x-1)=3\sin(1)+3\cos(1)\cdot(x-1)\approx 1.621x+0.904.$$ I'll leave $h$ for you to do.
Now, you've got (for some numbers $a$ and $b$): $$(1.621x+0.904)(ax+b)=1.621ax^2+(0.904a+1.621b)x+0.904b,$$ in which you don't care about the quadratic term (it's not accurate anyway).
Given that you only need a 1st-degree Taylor polynomial, it might be easier to just differentiate $f$ as written and compute it directly (not bother with $g$ and $h$ and multiplying).
The first-degree Taylor polynomial of $f(x)$ is the same as the tangent line through $(a, f(a))$, which is $y = f(a) + f'(a) (x - a)$.
Using the product rule, $f'(x) = 3 \cos x \cos x/2 - \frac{3}{2} \sin x \sin x/2$, hence $f'(1) \approx 0.817$. The tangent line passes through $(x,y) = (1, f(1))$, thus writing the line in the form $y = mx + c$, $3 \sin 1 \cos 1/2 = 0.817(1) + c \Rightarrow c \approx 1.398$. This matches the given answer of $y = 0.817x + 1.398$.
