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Find the gradient of g(x) = 4-x^2 at the point where x=2 by sketching a graph and finding the gradient of the tangent at x=2?
How do I do this by hand and by Cas? Thanks

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This is how I would find the gradient of g(x) at x=2:

Derivative: g'(x) = -2x
Evaluate derivative at x=2: g'(2) = -4
Gradient is -4

If you needed to do it graphically (not recommended since it is very imprecise!), you would:
1. Sketch 4-x^2 (as you would any parabola... you will notice it is (2-x)(2+x) and hence a negative parabola with x-intercepts at -2 and +2, and a y-intercept at y=4).
2. You would then take the point at x=2, and draw a tangent through it. The tangent is the line which touches x=2 on the parabola but doesn't go through the parabola (stays on the one side of it)
3. Then you'd measure the gradient of the tangent. You would hope it is close to -4, otherwise you didn't draw it accurately.

By CAS... you would follow the initial steps where I found the derivative, etc. except it would be a waste of time since it is so easy to do it by hand.