Equation of Tangent Line: Finding at a Given Point | Math HW Help

[teneleven]

Homework Statement

Find an equation of the tangent line to the curve at the given point.

y = 1 + 2x - x^3, (1,2)

Homework Equations

y = mx + b

m = \lim_{x\rightarrow a} \frac{f(x) - f(a)}{x - a}

The Attempt at a Solution

\lim_{x\rightarrow 1} \frac{(1 + 2x - x^3) - 2}{x - 1}

= \lim_{x\rightarrow 1} \frac{-1 + 2x - x^3}{x - 1}I'm not sure where to go from there.

 

[Spiro09]

your on the right track, try factoring the top line, the x-1 will cancel out

 

[Gib Z]

Rewrite the limit as \lim_{x\to 1} -\frac{x^3-2x+1}{x-1} then do some polynomial division.

 

[teneleven]

Thanks for the replies.

Factoring gets me...

\lim_{x\rightarrow 1} -x^2 + 1

Which results in "0"
Maybe I missed something. :/Using long division I get...

m = \lim_{x\rightarrow 1} x^2 + 2x - 1 = 1

\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = 1

The point given is (1,2) so...

y - 2 = 1(x - 1)

y = x + 1The answer in the book is y = -x + 3

I'm brushing up on my math in preparation for Calc III.

 

[Gib Z]

Try the long division again, you should have gotten it down to
m=\lim_{x\to 1} -(x^2+x-1) = -1
That leads to the answer in the book.

 

[HallsofIvy]

You could also, since this is "precalculus", use Pascal's method: If y= ax+ b is tangent to 1+ 2x- x³ at x= 1, then ax+ b- (1+ 2x- x³)= x³ +(a-2)x+ (b-1) has a double zero at x= 1. (Since the two graphs cross, clearly x= 1 makes that difference 0, since they are tangent there, it is a double 0.)

Then, setting x= 1, we have 1+ (a-2)+ (b-1)= a+ b- 2= 0 or b= 2- a. We can rewrite the equation as x³+ (a-2)x+ (1-a) and divide by x- 1 to get x²+ x+ (a-1) as quotient. The fact that x= 1 is a double zero of the first polynomial means it must make this 0 also. Taking x= 1 again, 1+ 1+ a-1= a+1= 0 so a= -1. b= 2-a= 2-(-1)= 3. The tangent line is y= -x+ 3.

Of course, it is far easier to note that the derivative of 1+ 2x- x³ is 2- 3x² and at x= 1, that is -1 so the tangent line is -x+ b and when x= 1, -1+ b= 2 so b= 3. But this is "precalculus"!
(You still have a ways to go to get to Calculus III.)

 

[Gib Z]

Calculus 3?! 2 courses of Calculus before you get to the derivative of polynomials? Your joking right?

 

[HallsofIvy]

teneleven did say, in his second post, "I'm brushing up on my math in preparation for Calc III."

Since this was posted in the Precalculus section, and he is taking the derivative by using the basic definition, I interpreted that to me that he is going back and reviewing all of the basics. That's why I said "You still have a ways to go".

 

[drpizza]

Another relevant equation to use might be:

f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}
Of course, it involves one more step to find f'(1)

 

[teneleven]

Great, thanks for your help. I solved it using the method in my original post.Now I'm trying to solve it using the equation drpizza pointed out.

f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}After plugging in y = -x^3 + 2x + 1 into the above equation I get...

\lim_{h\to 0} \frac{-(h^3 + 3h^2x + 3hx^2 +2x + 1)}{h}

After long division I'm left with...

\lim_{h\to 0} -(h^2 + 3x^2 +3hx) - \frac{2x + 1}{h}

I'm not sure what to do after this step. The point on the line I'm finding a tangent to is (1,2).

 

[HallsofIvy]

Great, thanks for your help. I solved it using the method in my original post.


Now I'm trying to solve it using the equation drpizza pointed out.

f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}


After plugging in y = -x^3 + 2x + 1 into the above equation I get...

\lim_{h\to 0} \frac{-(h^3 + 3h^2x + 3hx^2 +2x + 1)}{h}

No. f(x+h)= -(x+h)³+ 2(x+ h)+ 1= -x³-3hx²-3h²x- 1+ 2x+ 2h+ 1 so f(x+h)- f(x) is
-3hx²-3hx²+ h³+ 2h. All terms that do not involve h cancel out. (do you see why that must be true?) Then (f(x+h)- f(x))/h= -3x²- 3x²+ 2. What is the limit of that as h goes to 0?

After long division I'm left with...

\lim_{h\to 0} -(h^2 + 3x^2 +3hx) - \frac{2x + 1}{h}

I'm not sure what to do after this step. The point on the line I'm finding a tangent to is (1,2).