Find the largest and smallest values of $\mathit{y}\mathbf{=}\mathbf{4}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{9}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{12}\mathit{x}\mathbf{+}\mathbf{3}$ if $\mathit{x}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$.

It is given that, $y=4x^3+9x^2-12x+3 \dots \left(1\right)$.

Find the largest and the smallest value of (1), if $x=\cos \theta$.

To find the maximum and the minimum value of $y$, first differentiate$y$ with respect to$x$, and then equate to zero.

Calculating$\frac{dy}{dx}$ as:

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(4x^3+9x^2-12x+3\right) \\ \frac{dy}{dx}=4\frac{d}{dx}\left(x^3\right)+9\frac{d}{dx}\left(x^2\right)-12\frac{d}{dx}\left(x\right)+3\frac{d}{dx}\left(1\right) \\ \frac{dy}{dx}=4\left(3x^2\right)+9\left(2x\right)-12\left(1\right)+3\left(0\right) \\ \frac{dy}{dx}=12x^2+18x-12 \end{gathered}$$

Now equate the value of $\frac{dy}{dx}$equal to zero as:

$$\begin{gathered} \frac{dy}{dx}=0 \\ 12x^2+18x-12=0 \\ 2x^2+3x-2=0 \\ 2x^2+4x-x-2=0 \end{gathered}$$

Solving further as:

$$\begin{gathered} 2x\left(x+2\right)-1\left(x+2\right)=0 \\ \left(x+2\right)\left(2x-1\right)=0 \\ x=\frac{1}{2},-2 \end{gathered}$$

It can be observed that $x=\cos \theta$. This means that the value of $x$lies between $-1⩽x⩽1$. So, the value of $x$will be $\frac{1}{2}$.

Substitute the value of $x=\frac{1}{2}$in$y$ as:

$$\begin{gathered} y=4x^3+9x^2-12x+3 \\ y=4{\left(\frac{1}{2}\right)}^3+9{\left(\frac{1}{2}\right)}^2-12\left(\frac{1}{2}\right)+3 \\ y=\frac{1}{2}+\frac{9}{4}-3 \\ y=-\frac{1}{4} \end{gathered}$$

Equating$\frac{dy}{dx}=0$would find the maximum and minimum points for all the values of$x$as the function is continuous for all the values of $x$.

The range of the maxima and minima is$-1⩽x⩽1$. So, finding the value of$y$at$x=-1$and$x=1$.

Substitute $x=1$as:

$$\begin{gathered} y=4x^3+9x^2-12x+3 \\ y=4{\left(1\right)}^3+9{\left(1\right)}^2-12\left(1\right)+3 \\ y=4+9-12+3 \\ y=4 \end{gathered}$$

Now substitute$x=-1$ as:

$$\begin{gathered} y=4x^3+9x^2-12x+3 \\ y=4{\left(-1\right)}^3+9{\left(-1\right)}^2-12\left(-1\right)+3 \\ y=-4+9+12+3 \\ y=20 \end{gathered}$$

It can be observed that the value of $y$at the end point $x=-1$is greater than the value of $y$at $x=\frac{1}{2}$. This means that the minima exist at $x=\frac{1}{2}$and maxima exist at $x=-1$.

Therefore, the largest and smallest values of $y=4x^3+9x^2-12x+3$are$\left.\begin{array}{r}{y}_{\text{max}}=20\\ y_{\text{min}}=-\frac{1}{4}\end{array}\right\}x=\text{cos}\theta$.