AI HL · AHL 2
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Classifying stationary points and inflection
AHL 2.8
Find the stationary points of \(y=x^3+3x^2-24x\) and say whether each is a local maximum or a local minimum.
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dx}=3x^2+6x-24=3(x+4)(x-2)=0\)
A stationary point has gradient 0. Factorise to solve: x = −4 or x = 2.
\(x=-4:\ y=80\); \(x=2:\ y=-28\)
Put each x back into the original curve, not into the gradient.
x = −5: +; x = 0: −; x = 3: +
Test a value either side of each point in dy/dx. Gradient + to − is a maximum. Gradient − to + is a minimum.
(−4, 80) local maximum; (2, −28) local minimum
Give the coordinates and the type.
Find and classify the stationary points of \(y=x^4-4x^3\).
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dx}=4x^3-12x^2=4x^2(x-3)=0\), so x = 0 or x = 3
Factorise: the repeated factor gives x = 0.
\(\frac{d^2y}{dx^2}=12x^2-24x\)
At x = 3 it is 36 > 0: a local minimum, (3, −27).
\(\frac{d^2y}{dx^2}=0\)
The test does not decide. It could be a maximum, a minimum or a point of inflection. Check the gradient either side.
\(y'(-1)=-16\); \(y'(1)=-8\)
The gradient is negative on both sides, so it does not change sign. (0, 0) is a stationary point of inflection: the curve is not a maximum or a minimum.
Find the stationary point of \(y=x+4x^{-1/2}\) for \(x>0\), and classify it. Give coordinates to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(y'=1-2x^{-3/2}=0\), so \(x^{3/2}=2\)
Differentiate each term: 4x^(−1/2) gives −2x^(−3/2).
\(x=2^{2/3}=1.59\)
Raise both sides to the power 2/3.
\(y=4.76\)
Substitute into the original curve.
\(y''=3x^{-5/2}\), at x = 1.59: \(y''=0.945>0\)
Positive, so a local minimum at (1.59, 4.76).
Find the stationary point of \(y=e^{2x}-6x\) and classify it. Give coordinates to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(y'=2e^{2x}-6=0\), so \(e^{2x}=3\)
Divide by 2 first.
\(2x=\ln 3\), \(x=\tfrac12\ln 3=0.549\)
ln undoes e. Then divide by 2.
\(y=3-3\ln 3=−0.296\)
e^(2x) = 3 at this point, and 6x = 3 ln 3.
\(y''=4e^{2x}\) = 12 > 0
Positive, so a local minimum at (0.549, −0.296).
Find the stationary point of \(y=\ln x-\tfrac12x^2\) for \(x>0\), and classify it.
Have a go first. Then open the working one line at a time.
\(y'=\frac1x-x=0\)
Multiply by x: 1 = x², so x = 1. Reject x = −1: ln x needs x > 0.
\(y=\ln 1-\tfrac12=-0.5\)
Substitute x = 1 into the original curve.
\(y''=-\frac{1}{x^2}-1\), at x = 1: \(y''=-2<0\)
Negative, so a local maximum.
(1, −0.5) local maximum
State the coordinates and the type.
The function \(f(x)=2x^3+ax^2+bx-3\) has \(f''(-1)=-4\) and \(f'(2)=34\). Find the values of a and b.
Have a go first. Then open the working one line at a time.
\(f'(x)=6x^2+2ax+b\), \(f''(x)=12x+2a\)
Differentiate twice. a and b stay as letters.
\(-12+2a=-4\), so \(a=4\)
This equation has only a.
\(24+4a+b=34\), so \(24+16+b=34\)
Put in a = 4.
\(b=-6\)
Check: 24 + 16 − 6 = 34.
Find \(\frac{d^2y}{dx^2}\) for \(y=xe^{2x}\), and evaluate it at x = 0.
Have a go first. Then open the working one line at a time.
\(u=x,\ v=e^{2x}\), \(y'=1\times e^{2x}+x\times 2e^{2x}=(1+2x)e^{2x}\)
Product rule: u′v + uv′.
\(y''=2e^{2x}+(1+2x)\times 2e^{2x}=(4+4x)e^{2x}\)
Product rule again, with u = 1 + 2x.
\(y''(0)=4\)
Put x = 0: e⁰ = 1.
Sketch a curve for which, for all x, \(f(x)>0\), \(f'(x)<0\) and \(f''(x)>0\).
Have a go first. Then open the working one line at a time.
Every point is above the x-axis.
No x-intercepts.
The curve falls from left to right everywhere.
Gradient negative: decreasing.
The gradient is increasing, towards 0, so the curve is concave-up.
It falls more and more slowly.
\(y=e^{-x}\)
It is above the axis, always falling, and flattens out as x grows. Any curve with these three properties is correct.