Finding Tangent Line to a Curve at a Given Point

mbisCool
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Homework Statement



For what values of a and b is the line 3x+y=b tangent to the curve y=ax^3 when x=5 ?


The Attempt at a Solution


I believe i would need to find the equation of the tangent to y=ax^3 when x=5 then that should be equal to -3x+b=y if I am not mistaken but I am not sure what to do from here to solve for a and b.

Any leads in the right direction or insight if this is the entirely wrong approach would be greatly appreciated!
 
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I believe i would need to find the equation of the tangent to y=ax^3 when x=5 then that should be equal to -3x+b=y if I am not mistaken
Well, yes, that's just a restatement of the problem! You know that the slope of y= -3x+ b is -3. What connection is there between the slope of a tangent line and the function itself?
 
thank you for the quick reply hallsofivy. I had -0.04 instead of -3/75 giving me the wrong answer for the online homework webpage which was confusing me. After using -3/75 instead it counts it as correct :(
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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