# Maths - Calculus

#1

Graph the function. Describe briefly in words the interesting features of the graph including the location of critical points and where the function is increasing/decreasing. Then use derivative and algebra to explain the shape of the graph.

f(x) = x3 - 6x +1

The graph is increasing crossing the x-axis at about x=-2.5 and continuing positive into the second quadrant. The curve appears to peak around (-1.5,6.6) and then decreases crossing the y-axis around 1.2 continuing to fall. Around (1.4, -4.6) the curve begins and continues to increase.

f'(x) = 3x2 - 6

First we find the derivative, f'(x) = 3x2 - 6. Setting the derivative to 0, we have

3x2 - 6 = 0

3(x2 - 2) = 0

x2 - 2 = 0

x2 = 2

x = ±√2

critical points are x = √2 and x = -√2

Second we substitute for x in f'(x) = 3x2 - 6

f'(-5) = 69 +

f'(-4) = 42 +

f'(-3) = 21 +

f'(-2) = 6 +

f'(-√2) = 0

f'(-1) = -3 -

f'(0) = -6 -

f'(1) = -3 -

f'(√2) = 0

f'(2) = 6 +

f'(3) = 21 +

f'(4) = 42 +

f'(5) = 69 +

From this we see f(x) is increasing for all values where x<-√2 and all values where x > √2. F(x) is decreasing for all values -√2 < x < √2. This matches the original graph.

#2

Graph the function. Describe briefly in words the interesting features of the graph including the location of critical points and where the function is increasing/decreasing. Then use derivative and algebra to explain the shape of the graph.

f(x) = xe-x^2

The graph appears to begin around x=-2.5 and decreases to around (-.7,-.4) increasing through (0,0). At around (.7,.4) the curve begins decreasing and ends around x=2.5.

First we find the derivative using the product rule, (x)(dx)(e-x^2) + (e-x^2)(dx)(x)

f'(x) = x(-2x)(e-x^2) + (e-x^2)(1)

f'(x) = -2x2e-x^2 + e-x^2

Second we substitute for x in f'(x) = -2x2e-x^2 + e-x^2

f'(-3) = -0.0021 -

f'(-2) = -0.1282 -

f'(-1) = -0.3679 -

f'(-.71) = -0.0050 -

f'(-.7) = 0.0123 +

f'(-.3) = 0.7494 +

f'(0) = 1 +

f'(.3) = 0.7494 +

f'(.7) = 0.0123 +

f'(1) = -0.3679 -

f'(2) = -0.1282 -

f'(3) = -0.0021 -

#30

Let f(x) = x2 + cos(kx), for k>0

a) Using a calculator, graph f for k = 0.5, 1, 3, 5. Find the smallest number k at which you see points of inflection in the graph of k.

Around k=2.35 the points of inflection outside of the range -2 < x < 2 become difficult to make out. The points of inflection within that range become difficult to make out around k=1.45

b) Explain why the graph of f has no points of inflection if k<= √2, and infinitely many points of inflection if k>√2.

When k <= sqrt(2), the first derivative f'(x)=2x-(k*sin(kx)) only has one critical point where the graph crosses the x-axis (at x=0). The graph of the second derivative f''(x)=2-(k^2)(cos(x)) does not cross the x-axis at all and is defined everywhere, so there are no points of inflection.

When k > sqrt(2) however, f'(x) had three critical points, and f''(x) is a sinusoidal function crossing the x-axis at regular intervals. Each of these candidates for inflection points indeed change signs, thus we see an infinite number of inflections.

c) Explain why f has only a finite number of critical points , no matter what the value of k.

In a graph of f'(x) where for example k=5 : f'(x) = 2x - (5 * sin(5x)) we see that the sinusoid pattern of the sine function is displaced by the subtraction. Since the height of the sinusoid is fixed, the number of critical points where f'(x) can equal zero is limited by the length of time it takes for 2x to exceed the heights of that sine wave.

Product rule: (fg)' = f'g + fg'

Let f(x)=k and g(x)=sin(kx)

f'(x) = 0 since k in a constant

g'(x) = cos(kx) * k since it's derivative of sin(?) so we need a chain rule there.

(fg)' = 0*sin(kx) + k*(k(cos(kx))

(fg)' = (k2)(cos(kx))

So plugging that back into our original problem:

f''(x) = 2 - ( k2 * cos(kx) )

From this we see f(x) is decreasing for all values where -3< x > -.7, increasing for all values -.7 < x < 1 and then decreasing from 1< x > 3.

#4

What happens to concavity when functions are added?

a) If f(x) and g(x) are concave up for all x, is f(x) + g(x) concave up for all x?

Yes

b) If f(x) is concave up for all x and g(x) is concave down for all x, what can you say about the concavity of f(x) + g(x)? For example, what happens if f(x) and g(x) are both polynomials of degree 2?

c) If f(x) is concave up for all x and g(x) is concave down for all x, is it possible for f(x) + g(x) to change concavity infinitely often?

#5

Let p(x) = x3 -ax, where a is constant

a) If a<0, show that p(x) is always increasing.

substituting in values

b) If a>0, show that p(x) has a local maximum and a local minimum.

c) Sketch and label typical graphs for the cases when a<0 and when a>0.

Find the value(s) of x for which:

a) f(x) has a local maximum or local minimum. Indicate which ones are maxima and which are minima.

b) f(x) has a global maximum or global minimum

#6

f(x) = x10 - 10x, and 0 <= x <=2

#7

f(x) = sin2x - cos x, and 0 <= x <= π

#9

A line goes through the origin and a point on the curve y = x2e-3x, for x>=0. Find the maximum slope of such a line. At what x-value does it occur?

#### Solution Summary

Neat, step-by-step solutions are provided with neat graphs.