Prove that the max and min values of asinX + bcosX are respectively sqrt.(a^2+b^2) and -sqrt(a^2+b^2) .

1 answer:

6 0

The solution to the problem is as follows:
let y = asinx + bcosx

dy/dx = acosx - bsinx 

= 0 for max/min 

bsinx = acosx 

sinx/cosx = a/b 

tanx = a/b 

then the hypotenuse of the corresponding right-angled triangle is √(a^2 + b^2) 

the max/min of y occurs when tanx = a/b 

then sinx = a/√(a^2 + b^2) and cosx = b/√(a^2 + b^2) 

y = a( a/√(a^2 + b^2)) + b( b/√(a^2 + b^2)) 

= (a^2 + b^2)/√(a^2 + b^2) 

= √(a^2 + b^2)

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Simplify the expression (x^3- 5x^2 + 7x - 12) ÷ (x – 4) using long division

Kisachek [45]

Answer:

${x}^{2} - x + 3$

Step-by-step explanation:

Instead, since the divisor is in the form of $x - c$ , use what is called Synthetic Division. Remember, in this formula, $-c$ gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

4| 1 −5 7 −12

↓ 4 −4 12

_______________

1 −1 3 0 → ${x}^{2} - x + 3$

You start by placing the $c$ in the top left corner, then list all the coefficients of your dividend [x³ - 5x² + 7x - 12]. You bring down the original term closest to $c$ then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an ${x}^{2}$ , the −1 [ $x$ ] follows right behind it, and bringing up the rear, comes the 3, giving you the quotient of ${x}^{2} - x + 3$ .