MHB How Do You Find the Equation of a Tangent Line at a Given Point?

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To find the equation of the tangent line to the curve y = x^3 at the point P = (-3, -27), the slope (a) is determined by the derivative, which gives a = 27. Using the point-slope form of the line, the equation becomes y = 27x + b. Substituting the point P into the equation allows for the calculation of b, leading to the correct value of b being 54. Thus, the final equation of the tangent line is y = 27x + 54.
cgr4
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The line y = ax + b is tangent to y=x3 at the point P = (-3,-27). Find a and b.

I'm pretty lost on this one.

Here's my initial thoughts. Find (a) first.

(a) is equal to the slope which is the derivative of x3

So (a) would be equal to 3(-3)2 ?

Not quite sure where to start on this one.
 
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cgr4 said:
The line y = ax + b is tangent to y=x3 at the point P = (-3,-27). Find a and b.

(a) is equal to the slope which is the derivative of x3

So (a) would be equal to 3(-3)2 ?
Yes. Once you know $a$, $y=-27$ and $x=-3$, it is easy to find $b$.
 
Evgeny.Makarov said:
Yes. Once you know $a$, $y=-27$ and $x=-3$, it is easy to find $b$.

so a = 27

so, y = 27(-3) + b

-27 = -81 + b

therefore b = 64?
 
cgr4 said:
-27 = -81 + b

therefore b = 64?
Correct, except that 81 - 27 = 54.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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