AI SL · 12.9
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Maximum and minimum points on a domain
12.9.12
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(a) Find the local maximum point and the local minimum point on the curve \(y=x^3-3x^2-9x+10\) for \(-2\le x\le 6\). (b) Find the greatest value of y on this domain.
Have a go first. Then open the working one line at a time.
Set the window to \(-2\le x\le 6\)
The calculator only looks at the domain you give it.
From the calculator: local maximum (−1, 15), local minimum (3, −17)
Give both coordinates of each point.
x = −2 gives y = 8; x = 6 gives y = 64
The greatest value on a domain can be at an end-point.
Greatest value: y = 64 (when x = 6)
The end-point value is larger than the local maximum value.
Given that \(f(x)=x^4-6x^2+5\): (a) solve \(f(x)=0\); (b) find the coordinates of the points where the tangent is horizontal.
Have a go first. Then open the working one line at a time.
Graph \(y=x^4-6x^2+5\). From the calculator: zeros at x = −2.24, −1, 1 and 2.24
These are the x-intercepts of f.
Graph \(y=\frac{d}{dx}\left(x^4-6x^2+5\right)\). From the calculator: zeros at x = −1.73, 0 and 1.73
The tangent is horizontal where f′(x) = 0.
\(f(0)=5\), \(f(\pm1.73)=-4\)
Put each x into f(x), not into f′(x).
(0, 5), (−1.73, −4), (1.73, −4)
The lowest y-value is −4.
The curve \(y=x^4+bx^3+c\) has a horizontal tangent at the point (3, 10). Find the values of b and c.
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dx}=4x^3+3bx^2\)
The tangent is horizontal, so the gradient is 0 at x = 3.
\(4(27)+3b(9)=0\), so \(108+27b=0\) and b = −4
This equation has only b in it.
\(10=3^4+(-4)(3^3)+c=81-108+c\)
The point (3, 10) must fit the equation of the curve.
c = 37
Check: \(y=x^4-4x^3+37\) gives 10 at x = 3.
The curve \(y=ax^3+bx^2-12x\) has a horizontal tangent at the point (−1, 7). Find the values of a and b.
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dx}=3ax^2+2bx-12\); at x = −1: \(3a-2b-12=0\)
Horizontal tangent: the gradient is 0.
At (−1, 7): \(-a+b+12=7\), so \(-a+b=-5\)
Put x = −1 and y = 7 into the equation of the curve.
Solve \(3a-2b=12\) and \(-a+b=-5\) on the calculator: a = 2, b = −3
Two equations, two unknowns: use the solver.
\(\frac{dy}{dx}=6x^2-6x-12=0\) at x = −1
The gradient is 0 at x = −1, as required.
Let \(f(x)=2x^3-3x^2-12x+5\). (a) Find the values of x where \(f'(x)=0\). (b) Write down the interval on which f is decreasing. (c) State whether each stationary point is a local maximum or a local minimum.
Have a go first. Then open the working one line at a time.
\(f'(x)=6x^2-6x-12\). From the calculator: \(f'(x)=0\) at x = −1 and x = 2
Graph the derivative and read where it crosses the x-axis.
\(f'>0\) for x < −1, \(f'<0\) for −1 < x < 2, \(f'>0\) for x > 2
The graph of f′ is above the axis, then below, then above.
f is decreasing for \(-1<x<2\)
Where f′ is negative.
x = −1: f′ goes + to −, local maximum (−1, 12). x = 2: f′ goes − to +, local minimum (2, −15)
Positive then negative is a peak. Negative then positive is a valley.
The curve \(y=x^3-6x^2+12x+1\) has a horizontal tangent at the point (2, 9). Is (2, 9) a local maximum, a local minimum or neither?
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dx}=3x^2-12x+12=3(x-2)^2\)
At x = 2 the gradient is 0, as the question says.
x = 1: \(3(-1)^2=3\). x = 3: \(3(1)^2=3\)
Try one value on each side of x = 2.
The gradient is positive on both sides
A square is never negative, so \(3(x-2)^2\) never changes sign.
(2, 9) is neither a local maximum nor a local minimum
The curve flattens at x = 2 and keeps rising.