Question

f(x) = -(3/4)x2-8x3-42/5x2+105. Calculate local maxima and local minima.

Answer 1

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✴ $f(x) = \dfrac{ - 3}{4} {x}^{4} - {8x}^{3} - \dfrac{45}{2} {x}^{2} + 105$

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➡ The Point of local maxima is -5 & point of local minima is -3.

$\mathtt{\huge{\underline{\orange {Solution :-}}}}$

✭ $f(x) = \dfrac{ - 3}{4} {x}^{4} - {8x}^{3} - \dfrac{45}{2} {x}^{2} + 105}$

➝ $f'(x) = ( \frac{ - 3}{4} ) \times 4 {x}^{3} - 8 \times {3x}^{2} - ( \frac{45}{2} ) \times 2x$

➝ ${- 3x}^{3} - {24x}^{2} - 45$

➝ ${- 3(x ^{2} + 8x + 15)}$

➝ ${- 3( {x}^{2} + 5x + 3x + 15) }$

➝ ${- 3x(x + 5) + 3(x ^{2} + 5) }$

➝ $- 3(x + 5)(x + 3)$

Now, for the following critical point f'=0

⇾ $x = - 5 \: and \: x = - 3$

➩ Hence, For maxima is -5 & for minima is -3.

Answer 2

CORRECT QUESTION:

f(x) = - (3/4)x⁴ - 8x³ - 45/2x² + 105. Calculate local maxima and local minima.

ANSWER:

• Local maxima at x = 0, value = 105

• Local maxima at x = - 5, value = 73.75

• Local minima at x = - 3, value = 57.75

GIVEN:

• f(x) = - (3/4)x⁴ - 8x³ - 45/2x² + 105

TO FIND:

• Local maxima and local minima.

EXPLANATION:

$\sf f(x)= -\dfrac{3}{4}x^4 - 8x^3 - \dfrac{45}{2} x^2 + 105$

Differentiate w.r.t x

$\sf f(x)= -4\times\dfrac{3}{4}x^3 - 3(8)x^2 - 2\times\dfrac{45}{2} x$

$\sf f'(x)= -3x^3 - 24x^2 - 45 x$

$\sf f'(x)= -3x(x^2 + 8x+ 15 )$

By splitting the middle term

$\sf f'(x)= -3x(x^2 + 5x + 3x +15 )$

$\sf f'(x)= -3x(x(x + 5) + 3(x + 5))$

$\sf f'(x)= -3x(x+3)(x + 5)$

Put f'(x) = 0

$\sf 0= -3x(x+3)(x + 5)$

$\sf-3x = 0$

$\sf x = 0$

$\sf x + 3 = 0$

$\sf x = - 3$

$\sf x + 5 = 0$

$\sf x = -5$

$\sf f'(x)= -3x^3 - 24x^2 - 45 x$

Differentiate w.r.t x

$\sf f''(x)= -9x^2 - 48x - 45$

Substitute x = 0

$\sf f''(0)= -9(0)^2 - 48(0) - 45$

$\sf f''(0)= 0-0 - 45$

$\sf f''(0)= -45$

$\sf f''(0) < 0$

f(x) is maximum at x = 0

$\sf f''(x)= -9x^2 - 48x - 45$

Substitute x = - 5

$\sf f''(x)= -9(-5)^2 - 48(-5) - 45$

$\sf f''(x)= -9(25) +48(5) - 45$

$\sf f''(x)= -225 + 240  - 45$

$\sf f''(x)= 15  - 45$

$\sf f''(x)=-30$

$\sf f''(- 5) < 0$

f(x) is maximum at x = - 5

$\sf f''(x)= -9x^2 - 48x - 45$

Substitute x = - 3

$\sf f''(x)= -9(-3)^2 - 48(-3) - 45$

$\sf f''(x)= -9(9) +48(3) - 45$

$\sf f''(x)= -81 + 144  - 45$

$\sf f''(x)= -126+144$

$\sf f''(x)=18$

$\sf f''(- 3) > 0$

f(x) is minimum at x = - 3

Hence there are two local maximas at x = 0, x = - 5 and a local minima at x = - 3

$\sf f(x)= -\dfrac{3}{4}x^4 - 8x^3 - \dfrac{45}{2} x^2 + 105$

Substitute x = 0

$\sf f(0)= -\dfrac{3}{4}(0)^4 - 8(0)^3 - \dfrac{45}{2}( 0)^2 + 105$

$\sf f(0)= 105$

$\sf f(x)= -\dfrac{3}{4}x^4 - 8x^3 - \dfrac{45}{2} x^2 + 105$

Substitute x = - 5

$\sf f(-5)= -\dfrac{3}{4}(-5)^4 - 8(-5)^3 - \dfrac{45}{2}(-5)^2 + 105$

$\sf f(-5)= -\dfrac{3}{4}(625) - 8(-125) - \dfrac{45}{2}(25)+ 105$

$\sf f(-5)= -\dfrac{1875}{4}+ 1000 - \dfrac{1125}{2} + 105$

$\sf f(-5)= -468.75+ 1000 -562.5 + 105$

$\sf f(-5)= -1031.25+ 1000 + 105$

$\sf f(-5)= -31.25+ 105$

$\sf f(-5)= 73.75$

$\sf f(x)= -\dfrac{3}{4}x^4 - 8x^3 - \dfrac{45}{2} x^2 + 105$

Substitute x = - 3

$\sf f(-3)= -\dfrac{3}{4}(-3)^4 - 8(-3)^3 - \dfrac{45}{2}(-3)^2 + 105$

$\sf f(-3)= -\dfrac{3}{4}(81) - 8(-27)- \dfrac{45}{2}(9) + 105$

$\sf f(-3)= -\dfrac{243}{4}+216 - \dfrac{405}{2} + 105$

$\sf f(-3)= -60.75+ 216 -202.5 + 105$

$\sf f(-3)= -263.25+ 321$

$\sf f(-3)= 57.75$

Hence the two local maximas at x = 0, (value = 105), at x = - 5(value = 73.75) and a local minima at x = - 3(value = 57.75).