AI SL · 12.9
0 of 48 skills done
Stationary points: where the gradient is zero
12.9.11
Screenshot and video to come
Screenshot and video to come
Drag the orange point along the graph of f. Each place you stop leaves a dot on the lower graph at the height of the gradient. Trace f′ yourself, then press Show f′ to check.
Drag the point along f.
The curve \(y=ax^2+\frac{8}{x}\) has a horizontal tangent at x = 2. Find the value of a.
Have a go first. Then open the working one line at a time.
\(y=ax^2+8x^{-1}\)
Write the reciprocal as a power so you can use the power rule.
\(\frac{dy}{dx}=2ax-8x^{-2}=2ax-\frac{8}{x^2}\)
This is the gradient of the tangent at any x.
\(2a(2)-\frac{8}{4}=4a-2\)
You want the gradient at the point where x = 2.
\(4a-2=0\), so a = 0.5
A horizontal line has gradient 0.
If \(f(x)=x^3-9x^2+24x+5\), solve \(f'(x)=0\).
Have a go first. Then open the working one line at a time.
Graph \(y=\frac{d}{dx}\left(x^3-9x^2+24x+5\right)\)
From the calculator: the graph of f′(x), a parabola. Sketch it as evidence.
From the calculator: zeros at x = 2 and x = 4
The gradient is zero where the graph of f′ meets the x-axis.
\(f'(x)=3x^2-18x+24\); the quadratic solver gives x = 2 or 4
The derivative is a quadratic, so the solver agrees.
Solve \(f'(x)=0\) for \(f(x)=4x+\frac{25}{x}\), and give the positive solution.
Have a go first. Then open the working one line at a time.
Graph \(y=\frac{d}{dx}\left(4x+\frac{25}{x}\right)\)
From the calculator: f′ has a break at x = 0, because f is not defined there.
From the calculator: zeros at x = −2.5 and x = 2.5
f′(x) = 0 has two solutions.
\(f'(x)=4-\frac{25}{x^2}=0\), so \(x^2=\frac{25}{4}\)
The graph agrees with the algebra.
x = 2.5
The question asks for the positive solution.