Mathematics - Functions Question with Solution | TestHub

Given that $y=f(x)=3x^4-4x^3-12x^2+1$.

It cuts the y-axis at the point $(\mathrm{x}=0,\mathrm{x}=1)$.

Differentiating, we get $\frac{dy}{dx}=12x^3-12x^2-24x$

i.e. $\frac{dy}{dx}=12x\left(x^2-x-2\right)=12x(x-2)(x+1)$.

Now, $\frac{dy}{dx}=0\Rightarrow x(x-2)(x+1)=0\Rightarrow x=0,2,-1$

Also, $\frac{dy}{dx}>0\Rightarrow x(x-2)(x+1)>0$.

Using wavy-curve method, we have

Thus $\frac{dy}{dx}>0$ when $\mathrm{x}>2$ or $x\in (-1,0)$.

Similarly, $\frac{dy}{dx}<0$ when $0<\mathrm{x}<2$ or $\mathrm{x}<-1$.
Hence the graph of the curve will be as follows:

At $\mathrm{x}=2,\mathrm{f}(\mathrm{x})=3\times 16-4\times 8-12\times 4+1=48-32-48+1=-31$.

At $\mathrm{x}=-1,\mathrm{f}(\mathrm{x})=3.1+4.1-12.1+1=-4$.

∴ The least value of the function is -31 .

Hence the range of the function is $[-31,\infty )$.