Question

Find the largest and smallest values of $y=4x^3+9x^2-12x+3$ if $x=\cos \theta$.

Short answer

The largest value is 20 and the smallest value is -\frac{1}{4}.

Step-by-step solution

Express the function in terms of $\cos \theta$

Replace $x$ with $\cos \theta$ in the given function: $y=4(\cos \theta {)}^3+9(\cos \theta {)}^2-12(\cos \theta )+3$. Hence, we get the equation $y=4{\cos }^3\theta +9{\cos }^2\theta -12\cos \theta +3$.

Examine the range of $\cos \theta$

$\cos \theta$ ranges from -1 to 1, which is $[-1,1]$. So, we need to find the maximum and minimum values of $y$ in this interval.

Calculate the values at the endpoints

Evaluate the function at the endpoints of the interval. When $x=-1$: $$y=4(-1{)}^3+9(-1{)}^2-12(-1)+3=-4+9+12+3=20$$.When $x=1$: $$y=4(1{)}^3+9(1{)}^2-12(1)+3=4+9-12+3=4$$.

Check for critical points within the interval

To find any critical points, take the derivative of $y=4x^3+9x^2-12x+3$ with respect to $x$: $$\frac{dy}{dx}=12x^2+18x-12$$. Set the derivative equal to zero to solve for $x$: $$12x^2+18x-12=0$$. This can be simplified to $$2x^2+3x-2=0$$.Solve the quadratic equation for $x$: Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a=2$, $b=3$, and $c=-2$, we get $$x=\frac{-3\pm \sqrt{9+16}}{4}=\frac{-3\pm 5}{4}$$. Therefore, $x=\frac{1}{2}$ and $x=-2$ (but only $x=\frac{1}{2}$ lies within the interval $[-1,1]$).

Evaluate the function at the critical point

Evaluate the function at $x=\frac{1}{2}$: $$y=4{\left(\frac{1}{2}\right)}^3+9{\left(\frac{1}{2}\right)}^2-12\left(\frac{1}{2}\right)+3=\frac{4}{8}+\frac{9}{4}-6+3=\frac{1}{2}+\frac{9}{4}-6+3=\frac{2}{4}+\frac{9}{4}-6+3=\frac{11}{4}-6+3=\frac{11}{4}-\frac{24}{4}+\frac{12}{4}=\frac{11-24+12}{4}=\frac{-1}{4}$$.

Compare all values to find the maximum and minimum

The values obtained are $y=20$ at $x=-1$, $y=4$ at $x=1$, and $y=-\frac{1}{4}$ at $x=\frac{1}{2}$. Comparing these values, the largest value is 20 and the smallest value is -\frac{1}{4}.

Key concepts

Critical Points

To find the largest or smallest values of a function within a specific interval, it's essential to determine its critical points. Critical points occur where the derivative of the function is zero or undefined. First, derive the function:
$\frac{dy}{dx}=12x^2+18x-12$. Next, set the derivative equal to zero to identify critical points:
$12x^2+18x-12=0$. This simplifies to $2x^2+3x-2=0$. Solving this quadratic equation using the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, gives us:
$x=\frac{-3\pm \sqrt{9+16}}{4}$, resulting in $x=\frac{1}{2}$ and $x=-2$ (excluding $x=-2$ because it falls outside the interval $[-1,1]$).

Derivative

A derivative represents the rate of change of a function concerning its variable. In calculus, finding the derivative is key to understanding how the function behaves. For the given function:$y=4x^3+9x^2-12x+3$, we compute the first derivative to find critical points:
$\frac{dy}{dx}=12x^2+18x-12$. Setting this derivative equal to zero:
$12x^2+18x-12=0$. Solving this equation simplifies our search for the critical points, ensuring we evaluate the function accurately within its given range.

Quadratic Equation

Quadratic equations are polynomial equations of the form $a{x}^2+bx+c=0$. In our case, the simplified derivative results in:
$2x^2+3x-2=0$. Using the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=2$, $b=3$, and $c=-2$, we can find the roots:
$x=\frac{-3\pm 5}{4}$. This gives us the solutions $x=\frac{1}{2}$ and $x=-2$. The acceptable root within the interval $[-1,1]$ is $x=\frac{1}{2}$. Evaluating the function at these points helps pinpoint the minima and maxima needed for our optimization problem.

Trigonometric Functions

Trigonometric functions, such as cosine, play a critical role in many calculus problems. Given that $x=\cos \theta$ and $\cos \theta$ ranges within $[-1,1]$, we can rewrite the function as
$y=4{\cos }^3\theta +9{\cos }^2\theta -12\cos \theta +3$. By evaluating this equation at $x=-1$, $x=1$, and the critical point $x=\frac{1}{2}$, we note the values:
$y=20$ at $x=-1$,
$y=4$ at $x=1$, and $y=-\frac{1}{4}$ at $x=\frac{1}{2}$.
Comparing these, the maximum value is 20, and the minimum is $-\frac{1}{4}$.