Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

Consider the function, $F (x) = \int_{- 1}^{x} (t^{2} - t) d t, x \in R$ . Find

(a) The $x$ and $y$ intercept of $F$ if they exist.

(b) Derivatives $F' (x)$ and $F'' (x)$ .

(c) The interval on which $F$ is an increasing and the intervals on which $F$ is decreasing.

(d) Relative maximum and minimum points.

(e) Any inflection point.

Accepted Answer

Given the function, $F (x) = \int_{- 1}^{x} (t^{2} - t) d t, x \in R$ .

$\Rightarrow F (x) = {[\frac{t^{3}}{3} - \frac{t^{2}}{2}]}_{- 1}^{x}$

$\Rightarrow F (x) = [\frac{x^{3}}{3} - \frac{x^{2}}{2} - (- \frac{1}{3} - \frac{1}{2})]$

$\Rightarrow F (x) = \frac{x^{3}}{3} - \frac{x^{2}}{2} + \frac{5}{6} .$

(a) To find $x$ -intercept of $F$ , put $y = F (x) = 0 .$

$\Rightarrow \frac{x^{3}}{3} - \frac{x^{2}}{2} + \frac{5}{6} = 0$

$\Rightarrow 2 x^{3} - 3 x^{2} + 5 = 0$

$\Rightarrow (x + 1) (2 x^{2} - 5 x + 5) = 0$

$\Rightarrow x = - 1$ or $2 x^{2} - 5 x + 5 = 0,$ the discriminant of $2 x^{2} - 5 x + 5 = 0$ is ${(- 5)}^{2} - 4 \times 2 \times 5 < 0,$ hence no real roots.

Thus, the $x$ -intercept is $(- 1, 0) .$

Now to find the $y$ -intercept, put $x = 0, \Rightarrow y = F (0) = \frac{5}{6} .$

Thus, the $y$ -intercept is $(0, \frac{5}{6}) .$

(b) Derivatives $F^{'} (x) = \frac{d}{d x} (\frac{x^{3}}{3} - \frac{x^{2}}{2} + \frac{5}{6}) = x^{2} - x$ and $F^{''} (x) = \frac{d}{d x} (x^{2} - x) = 2 x - 1 .$

(c) We have $F^{'} (x) = x (x - 1)$ .

And $x (x - 1) > 0$ for $(- \infty, 0) \cup (1, \infty)$ and $x (x - 1) < 0$ for $(0, 1)$ .

Thus, the interval in which $F$ is an increasing is $(- \infty, 0) \cup (1, \infty)$ and the interval in which $F$ is decreasing is $(0, 1) .$

(d) We have $F^{'} (x) = x^{2} - x = x (x - 1)$ and to find the point of relative maxima or minima, put $F^{'} (x) = 0 .$

$\Rightarrow x (x - 1) = 0$

$\Rightarrow x = 0$ or $x = 1 .$

Also, $F^{''} (x) = 2 x - 1,$ and at $x = 0, F^{''} (0) = - 1 < 0$ and at $x = 1, F^{''} (1) = 2 - 1 = 1 > 0 .$

Hence, $x = 0$ is the point of relative maximum and $x = 1$ is the point of relative minimum.

Also, at $x = 0, y = \frac{5}{6}$ and at $x = 1, y = \frac{1}{3} - \frac{1}{2} + \frac{5}{6} = \frac{2}{3} .$

Thus, the relative maximum point is $(0, \frac{5}{6})$ and relative minimum point is $(1, \frac{2}{3}) .$

(e) For the inflection point $F^{''} (x) = 0 .$

$\Rightarrow 2 x - 1 = 0$

$\Rightarrow x = \frac{1}{2} .$

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Consider the function, Fx=∫-1xt2-tdt, x∈R. Find (a) The x and y intercept of F if they exist. (b) Derivatives F'x and F''x. (c) The interval on which F is an increasing and the intervals on which F is decreasing. (d) Relative maximum and minimum points. (e) Any inflection point.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !