Determine the Horizontal tangents of f(x)

Question

What are the steps used to find horizontal tangents of a function f(x) as defined:
$$ f(x)=(x+1)^2 (2x-3)^3 $$

I know we have to calculate the derivative of the function, but how do we do that? (Only looking for the real roots of the derivation = 0)

After using the Product rule $u=(x+1)^2$ and $v=(2x-3)^3$ i get a $(uv)'=uv'+u'v = 40x^4-80x^3-30x^2+90x$, but how do we find when this derivitive becomes 0 ?

Answer 1

$$\frac{d}{dx}( (x+1)^2 (2x-3)^3)=0$$

Some formulae $$\text{i] }\frac{d}{dx} (x^n) = n x^{n-1}$$ $$\text{ii] }\frac{d}{dx}( (ax+b)^n)= a n (ax+b)^{n-1} $$

As $$\frac {d}{dx} (f(g(x)) = f'(g(x)). g'(x) $$ $$\text{where } f'(x)=\frac{d}{dx} ( f(x))$$

And $$\frac{d}{dx} (uv) = u \frac {d}{dx} (v) + \frac {d}{dx} (u) v $$

$$(x+1)^2 \frac {d}{dx} ((2x-3)^3)+(2x-3)^3 \frac{d}{dx} ((x+1)^2)=0$$

$$ \text{.:}(x+1)^2 3 (2x-3)^2 2 + (2x-3)^3 2(x+1) =0$$ $$ \text{.:} (2x-3)^2(x+1)(6(x+1)+2(2x-3))=0$$ $$ \text{.:}(2x-3)^2 (x+1) (6x +6+4x-6)=0$$ $$ \text{.:}(2x-3)^2 (x+1) (10x) =0$$ $$2x=3 \text{ , }x=-1 \text{ , } x=0$$ $$x=\frac{3}{2} \text{ , } x=-1 \text{ , } x=0$$

Answer 2

HINT: use that $$f'(x)=10\, \left( x+1 \right) \left( 2\,x-3 \right) ^{2}x$$