Let f(x)=3^x-x^3. The tangent to the curve is parallel to the secant through (0,1) and (3,0) for x ~? how do you go about to solve this?
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Originally Posted by got_jane Let f(x)=3^x-x^3. The tangent to the curve is parallel to the secant through (0,1) and (3,0) for x ~? how do you go about to solve this? The gradient of the line passing through (0,1) and (3,0) is 1/3. So solve f'(x) = 1/3 for x. Exact solutions are probably not possible .....
Originally Posted by mr fantastic The gradient of the line passing through (0,1) and (3,0) is 1/3. So solve f'(x) = 1/3 for x. Exact solutions are probably not possible ..... so what can I do to approximate?
Originally Posted by got_jane so what can I do to approximate? You use whatever technology you have access to (graphics calculator, CAS calculator, software such as mathematica, derive etc.) in the appropriate way.
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