MATEMATİK I

If f(x)=2x³-3x²+4x-3, then f'(x)=6x²-6x+4. And for x=2 f'(x)=16

The average velocity is defined by v_av=(f(t₂)-f(t₁))/(t₂-t₁). So we have to calculate f(t) for t=1 and for t=3.

f(t)=5t²-2t+1

f(3)=45-6+1=40

f(1)=5-2+1=4

Then, v_av=(f(t₂)-f(t₁))/(t₂-t₁)=(40-4)/(3-1)=18 for interval [1,3].

If a function y = f(x) is differentiable, its derivative f '(x) is a function of x.

If the derivative f '(x) is differentiable, the derivative of f '(x) is called the second derivative of y = f(x) and is denoted by y˝=f ˝(x).

The derivative of the second derivative, i.e., f ˝(x) is called the third derivative which is denoted by y'''= f ''' (x). 

For a constant function f(x) = c the derivative is zero, that is, (c)' = 0. 

The slope of the tangent line to a curve y = f(x) at (x0, f (x0 )) is m = f ' (x0). 

For the identity function f(x) = x, the derivative is 1, that is, (x)' = 1. The answer is A.

The derivative of the function f is equal to 3x² - 2. When the point is plug into the derivative it yields to 3*(0)² - 2 = -2. The answer is -2, choice A.

For the car given, t₁=1 and t₂=5. The values of f at these points are f(t₁) = f(1) = 1³+1²+1 = 3,
f(t₂) = f(5) = 5³ + 5² +5 = 155. So, the average velocity is equal to (155 - 3)/(5 - 1) = 152/4 = 38 metres/seconds. The answer is A.

In order to find the instantenous velocity the derivative of the function must be taken. This is equal to 3t² + 2t + 1. The velocity at this instant is equal to the value found when t=2 is plugged in: (3*2²)+ (2*2) + 1 = 12 + 4 + 1= 17(metre/seconds). The answer is B.

The derivative of  f(x) = 2x³ + 5x + 3 is equal to f'(x) = 6x² + 5. At point x=3, f'(3) = 6*(3)² + 5 = 54 + 5 = 59. The answer is C.

The derivative of the function f(x) = 2x³ + 3x² + 5x is equal to 6x² + 6x + 5. f'(2) = 6*(2)² + 6*2 + 5 = 24+12+5=41 and f(1) = 2*(1)³ + 3*(1)² + 5*1 = 2+3+5=10. the sum of these two values 41+10= 51. The answer is C.

The Chain Rule: The derivative of composite function f (g(x)) is f'(g(x)) * g' (x). For this question our g(x) = 2x+1. The derivative of (2x + 1)³ yields to 3(2x + 1)² * 2 which is equal to 6(2x + 1)² . The answer is D.

For the function  x³ + 2x² - 4x the derivative is equal to 3x² + 4x - 4 and the second derivative is equal to 6x + 4. If we equal the derivative functions to zero we would have  3x² + 4x - 4=0 is (3x-2)(x+2)=0 x will be equal to -2 and 2/3, these are the critical points. When we plug each one to the second derivative we will find 6(-2)+4=-8 and 6(2/3)+4=8. -8<0 which makes x=-2 a local maximum and 8>0 which makes x=2/3 a local minimum. The answer is D.

The monotonicity property of a function can be investigated by the sign of the derivative. If the
inequality f '(x) > 0 is satisfied for all x ∈ (a, b) then f(x) is called monotone increasing on the
interval (a, b); likewise, if f '(x) < 0 for all x ∈ (a, b) then f(x) is called monotone decreasing on
the interval (a, b). The answer is C.

Second derivative test for local extrema: if a is a critical point and the second derivative f ˝(a) is
positive, i.e. f ˝(a) > 0, then a is a local minimum point. If f ˝(a) is negative, i.e. f ˝(a) < 0, then a
is a local maximum point. For f(x) the derivative is equal to f'(x)=4x+2 and the second derivative is equal to f''(x)=4 this value is a positive value which indicates a local minimum point. If we equal the derivative to zero, f'(x)=4x+2=0 the x value would be equal to -1/2. The answer is E.

f(x)=x²+2x+lnx

f'(x)=2x+2+(1/x) and for x=1 f'(1)=2+2+1=5

The first derivative must be equal to zero and the second derivative must be positive for a minimum value of a function. Thus:

f(x)=x²-6x+5

f'(x)=2x-6=0 so 2x=6, x=3 and when x=3 y=f(x)=9-18+5=-4

The second derivative is:

f''(x)=2>0

Thus for given function, point (3,-4) is the minimum.

f(x)=x²+lnx-1

f'(x)=2x+(1/x)

f''(x)=2-(1/x²)

Let x and y be the side lengths of the rectangele. If the perimeter is 80, then 2x+2y=80

x+y=40

y=40-x

The are of the rectangle is A=x*y=x*(40-x)=40x-x²

Since it is a maximization problem the first derivative of the area function (A) must be equal to zero and the second derivative must be negative.

A''(x)=-2<0 (Second derivative rule satisfied)

A'(x)=40-2x=0, 40=2x, 20=x, y=40-x=20 So the area A=20*20=400

f(g(x))=(2x+1)^1/2

From the chain rule we know that (f(g(x)))'=f'(g(x))g'(x)

Applying the chain rule we get (f(g(x)))'=(1/2)(2x+1)^-1/22=(2x+1)^-1/2

The equation of tangent line is given by the formula:

y – f(x₀) = f'(x₀)(x – x₀)

x₀=2 and f(x₀)=1 are given.

Since f(x)=x²-2x+1 then f'(x)=2x-2 and f'(x₀)=2*2-2=2 for x₀=2.

Thus the equation for the tangent line will be:

y-1=2(x-2)

y-1=2x-4

y=2x-3

For the extreme points the first derivative must be equal to zero. Since f(x)=x³-4x The first derivative of function f(x) will be:

f'(x)=3x²-4

Then we find the roots of the f'(x) as following:

3x²-4=0

3x²=4

x= -√ 2/3 and x= √ 2/3 are the roots of this equation. So the answer is D.

A function is increasing where its first derivative is positive. Thus, we have to find the intervals where the first derivative of function is positive.

f(x)=x³-12x

f'(x)=3x²-12

So the function is increasing where f'(x)>0

Namely:

3x²-12>0

3x²>12

x²>4

This is possible when x>2 and when x<-2

Thus the function is increasing in the interval (-∞, -2)U(2,∞)