Chapter 4.4, Problem 1E

To determine

Tofind:the interval of increase and decrease function, interval of concavity, local maxima,local minima and inflection points using graph.

Answer to Problem 1E

From the graph it can be easily verify the increasing and decreasing of the function, maxima and minima and inflection point and concavity of the function.

Explanation of Solution

Given:

  $\text{f}\left(\text{x}\right){\text{=4x}}^{\text{4}}{\text{-32x}}^{\text{3}}{\text{+89x}}^{\text{2}}\text{-95x+29}$ .

Concept used:

Increasing or decreasing function can be calculated by equating first derivative of the function to 0.

  $\frac{\text{dy}}{\text{dx}}\text{=0}$ .

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{>0}\text{.}$ the concave will be open upward and local minima can be found.

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{<0}\text{.}$ the concave will open downward and local maxima can be found.

If the graph of $\text{y=<mo>f</mo>}\left(\text{x}\right)$ has point of inflection then .

Calculation:

  $\text{f}\left(\text{x}\right){\text{=4x}}^{\text{4}}{\text{-32x}}^{\text{3}}{\text{+89x}}^{\text{2}}\text{-95x+29}$ .

  

  $\text{16}{\left(\text{x-2}\right)}^{\text{3}}\text{-14}\left(\text{x-2}\right)\text{+5=0}$ .

  $\text{2x}\left({\text{8x}}^{\text{2}}\text{+89}\right){\text{=96x}}^{\text{2}}\text{+95}$ .

  $\left(\text{2x-5}\right)\left({\text{8x}}^{\text{2}}\text{-28x+19}\right)\text{=0}$ .

  $x=\frac{5}{2}$ .

  $\text{x=}\frac{\text{7}}{\text{4}}\text{-}\frac{\sqrt{\text{11}}}{\text{4}}$ .

  $\text{x=}\frac{\text{7}}{\text{4}}+\frac{\sqrt{\text{11}}}{\text{4}}$ .

The function increases in the interval of

  $\left(\frac{\text{7}}{\text{4}}\text{-}\frac{\sqrt{\text{11}}}{\text{4}},\frac{\text{7}}{\text{4}}+\frac{\sqrt{\text{11}}}{\text{4}}\right),\left(\frac{\text{7}}{\text{4}}+\frac{\sqrt{\text{11}}}{\text{4}},\frac{5}{2}\right),\left(\frac{5}{2},\infty \right)$ .

The function decreases in the interval of

  $\left(-\infty ,\frac{\text{7}}{\text{4}}\text{-}\frac{\sqrt{\text{11}}}{\text{4}}\right)$ .

  

   .

  $\text{48}\left(\text{x-4}\right)\text{x+178=0}$ .

  $\frac{\text{24}}{\text{7}}{\left(\text{x-2}\right)}^{\text{2}}\text{=1}$ .

  $\text{48}{\left(\text{x-2}\right)}^{\text{2}}\text{-14=0}$ .

  $\text{x=2-}\frac{\sqrt{\frac{\text{7}}{\text{6}}}}{\text{2}}$ .

  $\text{x=2+}\frac{\sqrt{\frac{\text{7}}{\text{6}}}}{\text{2}}$ .

Concave upward at point which is Minima $x=2.54$ , $\text{y=-0}\text{.041}$ .

Concave downward at point which is maxima $\text{x=1}\text{.46}$ , $\text{y=10}\text{.041}$ .

Inflection point lies in the interval of:

  $\text{f}\left(\text{x}\right)\text{=}\frac{{\text{x}}^{\text{2}}\text{-1}}{{\text{x}}^{\text{3}}}$ where $\text{x=2}\pm \frac{\sqrt{\frac{\text{7}}{\text{6}}}}{\text{2}}$ .

Put the value of x here to get the value of y coordinate:

  $f\left(\text{2-}\frac{\sqrt{\frac{\text{7}}{\text{6}}}}{\text{2}}\right)=10.041$

  $f\left(\text{2+}\frac{\sqrt{\frac{\text{7}}{\text{6}}}}{\text{2}}\right)=-0.041$ .

Inflection point= $\text{y=10}\text{.041, y=-0}\text{.041}$ .

The function increases in the interval of $\left(-\sqrt{3},0\right),\left(0,\sqrt{3}\right)$

The function decreases in the interval of $\left(-\infty ,-\sqrt{3}\right),\left(\sqrt{3},\infty \right)$ .

Concave upward at point which is Minima $x=2.54$ , $\text{y=-0}\text{.041}$ .

Concave downward at point which is maxima $\text{x=1}\text{.46}$ , $\text{y=10}\text{.041}$ .

Inflection point= $\text{y=10}\text{.041, y=-0}\text{.041}$ .

The graph of the function

  

Red line indicates the graph of the function $\text{f}\left(\text{x}\right)$ .

Blue line indicates the graph of the first derivative of the function .

Green line indicates the second derivative of the function .

Hence from the graph it can be easily verify the increasing and decreasing of the function, maxima and minima and inflection point and concavity of the function.