7.1 & 7.3 Sketching Graphs

What is a stationary point?

A point on a graph with a gradient of 0. The graph has flattened out. A local maximum / minimum / point of inflection

How do you sketch a graph?

1) Find where the graph crosses the axes by substituting in y=0 to find where it crosses the x-axis and x=0 for the y-intercept 2) Differentiate the function of the graph 3) Substitute in f'(x)= 0 to find the stationary points and calculate the y values that these happen at (sub in values of x into function) 4) Use f''(x) to determine the nature of the stationary point] 5) Factorise out the highest power of x and see what happens as x gets really big or smaller to find general shape (e.g. factorise out x² if quadratic)

Find the interval in which f(x) = x³ + 3x² - 9x is decreasing

Decreasing when f'(x) < 0 f(x) = x³ + 3x² - 9x f'(x) = 3x² + 6x - 9 3x² + 6x - 9 < 0 x² + 2x - 3 < 0 (x + 3)(x - 1) < 0 -3 < x < 1 so f(x) is decreasing in the interval [-3, 1]

If f''(a) when x = a is 0, what type of stationary point is at a?

Don't know as could be a maximum, minimum or point of infection Must look at the gradient either side of the stationary point (f'(x + h) and f'(x - h))

How can you determine the nature of a turning point?

Find the second derivative of curve at that point to see how the gradient is changing

If f''(a) when x = a is less than 0, what type of stationary point is at a?

Local maximum as the gradient is decreasing from positive to negative gradient

If f''(a) when x = a os greater than 0, what type of stationary point is at a?

Local minimum as the gradient is increasing from negative to positive gradient

Show that f(x) = 4 - x(2x² + 3) is decreasing for all x∈R

Real numbers can represent a distance on a line (not imaginary) f(x) = 4 - x(2x² + 3) f(x) = 4 - 2x³ - 3x f'(x) = -6x² - 3 x² ≥0 for all x∈R so -6x² - 3 is < 0 for all x∈R

When is a function decreasing?

The function f(x) is (strictly) decreasing in the interval [a, b], if f'(x) < 0 for values of x between a and b (the gradient is negative so values of y are decreasing as values of x increase)

When is a function increasing?

The function f(x) is (strictly) increasing in the interval [a, b], if f'(x) > 0 for values of x between a and b (the gradient is positive so values of y are increasing as values of x increase)

What is the second (order) derivative?

The rate of change of the gradient function / f''(x) / d²y/dx²

What is a point of inflection?

f''(x) = 0 as the rate of change of gradient changes from increasing to decreasing or vice versa (concave to convex) f''(x) changes from > 0 to f''(x) < 0 or vice versa (the gradient is increasing and then decreasing). However, the gradient changes either side but stays the same sign (e.g. negative and then negative so always decreasing function but gradient first decreasing then increasing)

What is the gradient like either side of a local minimum on the curve f(x)?

f'(a - h) = negative f'(a) = 0 f'(a + h) = positive

What is the gradient like either side of a local maximum on the curve f(x)?

f'(a - h) = positive f'(a) = 0 f'(a + h) = negative

What is the gradient like either side of a point of inflection on the curve f(x)?

f'(a - h) = positive / negative f'(a) = 0 f'(a + h) = positive / negative

f(x) = x⁴ + 3x³ - 5x² - 3x + 1 Find the coordinates of stationary points, the nature of each and sketch the graph

f(x) = x⁴ + 3x³ - 5x² - 3x + 1 f'(x) = 4x³ + 3x³ - 10x - 3 f''(x) = 12x² + 18x - 10 Turning points at: 4x³ + 3x³ - 10x - 3 = 0 f'(1) = 0 so (x - 1) is a factor ......... .........____________________ x - 1 |4x³ + 3x³ - 10x - 3 .......- (4x³ - 4x²)................(4x²(x - 1)) ...................13x² - 10x ................- (13x² - 13x).....(13x(x - 1)) ................................3x - 3 .............................- (3x - 3)..(3(x - 1)) ........................................0 (x - 1)(4x² + 13x + 3) = 0 (x - 1)(4x + 1)(x + 3) = 0 x = 1, x = -¼, x = -3 y = -3, y = 356/256, y = -35 f''(1) = 20 > 0 so (1, -3) is a minimum f''(-¼) = -55/4 < 0 so (-¼, 357/256) is a maximum f''(-3) = 44 > 0 so (-3, -35) is a minimum Graph is a W-shaped quartic with three turning points (two minimums below the x-axis and a central maximum above the x-axis)

Find the coordinates of the stationary points on the curve y = 2x³ - 15x² + 24x + 6. Determine the nature of the stationary points

y = 2x³ - 15x² + 24x + 6 dy/dx = 6x² - 30x + 24 d²y/dx² = 12x - 30 6x² - 30x + 24 = 0 x² - 5x + 4 = 0 (x - 4)(x - 1) = 0 x = 4, x = 1 y = 2(4)³ - 15(4)² + 24(4) + 6 y = -10 y = 2(1)³ - 15(1)² + 24(1) + 6 y = 17 When x =4, 12(4) - 30 = 18 so (4, -10) = max When x=1, 12(1) - 30 = -18 so (1, 17) = min