Chapter 2.3, Problem 23E

Solution

To Sketch: The given function with local and minimum extrema and $x-$ intercepts. And explain its end behaviour with upper and lower limits.

Given: $f\left(x\right)=2x^4-5x^3-17x^2+14x+41$

Graph:

Sketch the graph of $f\left(x\right)=2x^4-5x^3-17x^2+14x+41$with all its extrema and zeros.

  

Find the extrema:

Find the first derivative of the function:

  $f\left(x\right)=8x^3-15x^2-34x+14$

Find the second derivative of the function:

  $f^{\prime \prime }\left(x\right)=24x^2-30x-34$

To find the local maximum and minimum values of the function, set the derivative equal to  $0$  and solve.

  $8x^3-15x^2-34x+14=0$

Graph each side of the equation. The solution is the $x-$ value of the point of intersection.

  $x\approx -1.56230867,0.36453633,3.07277234$

Evaluate the second derivative at $x=-1.56230867$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

  $\begin{array}{l}$ =24{(-1.56230867)}^2-30\cdot -1.56230867-34$=\text{71}\text{.44866169}\end{array}$

  $x=-1.56230867$ is a local minimum because the value of the second derivative is positive.

Evaluate the second derivative at $x=0.36453633$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

  $\begin{array}{l}$ =24{(0.36453633)}^2-30\cdot 0.36453633-34$=-41.74680826\end{array}$

  $x=0.36453633$ is a local maximum because the value of the second derivative is negative.

Evaluate the second derivative at $x=3.07277234$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

  $\begin{array}{l}$ =24{(3.07277234)}^2-30\cdot 3.07277234-34$=100.42314656\end{array}$

  $x=3.07277234$ is a local minimum because the value of the second derivative is positive.

The function $f$ has a local minimum of about $-86.497$ at $x=\frac{5}{8}$ .

Find the values of $y:$

Substitute $x=-1.56230867$ in $f\left(x\right)=2x^4-5x^3-17x^2+14x+41$ : $\begin{array}{l}y=2{\left(-1.56230867\right)}^4-5{\left(-1.56230867\right)}^3-17{\left(-1.56230867\right)}^2+14\left(-1.56230867\right)+41\\ y=8.61550772\end{array}$

Substitute $x=0.36453633$ in $f\left(x\right)=2x^4-5x^3-17x^2+14x+41$:

  $\begin{array}{l}y=2{\left(0.36453633\right)}^4-5{\left(0.36453633\right)}^3-17{\left(0.36453633\right)}^2+14\left(0.36453633\right)+41\\ y=43.63754166\end{array}$

Substitute $x=3.07277234$ in $f\left(x\right)=2x^4-5x^3-17x^2+14x+41$:

  $\begin{array}{l}y=2{\left(3.07277234\right)}^4-5{\left(3.07277234\right)}^3-17{\left(3.07277234\right)}^2+14\left(3.07277234\right)+41\\ y=-43.25842048\end{array}$

The function $f$ has 3 extrema of about $-43.258$ at $x\approx 3.072$ and of about $8.615$ at $x\approx -1.562$ and of about $43.637$ at $x\approx 0.364$ .

Find the zeroes of the function $f\left(x\right)=2x^4-5x^3-17x^2+14x+41$:

  $2x^4-5x^3-17x^2+14x+41=0$

Graph each side of the equation. The solution is the x-value of the point of intersection. The function has zeros at $x\approx 1.85980232\left(\text{twice}\right),3.87504384\left(\text{twice}\right)$

Thus, the function $f$ has $3$ extrema and $4$ zeros, the maximum number possible for a quadratic.

End behaviour:

Use the limits to describe its end behaviour.

  $\underset{x\to \infty }{\lim }f\left(x\right)=\infty \text{ and }\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty$

The graph rises to the left and rises to the right.