Find the maximum value of ${(7 - x)}^{4} {(2 + x)}^{5}$ when $x$ lies between $7$ and $- 2$ .

Open in App

Solution

Points where derivative of the function is equal to zero are either maxima or minima since it is a continuous function.
putting derivative = 0 , we get
$(4 \times - 1 \times {7 - x}^{3} \times {2 + x}^{5}) + (5 \times {7 - x}^{4} {2 + x}^{4}) = 0$ .
or , $5 {7 - x} = 4 {2 + x}$ since $x \in (- 2, 7)$ so ${x + 2} \neq 0 & {7 - x} \neq 0$ .

Therefore $9 x = 27$ or , $x = 3$ .
Hence the maximum value will be ${(7 - 3)}^{4} {(2 + 3)}^{5}$ or $8 \times 10^{5}$ .

Suggest Corrections

Similar questions

x^{5} - x^{4} + x^{3} - x^{2} + 1

x = 2

You are given $c o s x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .;$

$s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$

$t a n x = x + \frac{x^{3}}{3} + \frac{2. x^{5}}{15} . . . . . .$

Find the value of $lim x \to 0 \frac{x c o s x + s i n x}{x^{2} + t a n x}$

You are given $c o s x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .;$

$s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$

$t a n x = x + \frac{x^{3}}{3} + \frac{2. x^{5}}{15} . . . . . .$

Find the value of $lim x \to 0 \frac{x c o s x + s i n x}{x^{2} + t a n x}$

x

- 6

8

(x + 6)^{4} (8 - x)^{3}

$a = 1 + \frac{x^{3}}{3!} + \frac{x^{6}}{6!} + . . .$ ,

$b = x + \frac{x^{4}}{4!} + \frac{x^{7}}{7!} + . . .$ ,

$c = \frac{x^{2}}{2!} + \frac{x^{5}}{5!} + \frac{x^{8}}{8!} + . . .$ .

If $a^{3} + b^{3} + c^{3} - z a b c = 1$ . Find the value of $z$

Join BYJU'S Learning Program