# Finding the Equation of a Tangent Line and Limits for a Curve

• squenshl
In summary: Hold up.I can just calculate ##\int_{0.5}^{1} (2x-1)^4 \; dx=0.1##.Then calculate ##\int_{7/8}^{1} 8x-7 \; dx=0.0625##.Now take the difference to get...0.5862
squenshl
Member warned to type the problem statement, not just post an image with type that is too small to read

See attached.

## The Attempt at a Solution

Ok so the first thing you want to do is find the equation of the tangent line which is done in the usual way to get ##y=8x-7##. This cuts the ##x## axis at ##x=\frac{7}{8}## so we know the upper limit of the integral to calculate.
Is ##P## a turning point if not how do I find it?

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squenshl said:
See attached.
Can't read the very small print. write out a problem statement, sort out what you are supposed to do !

I"m having the same problem. When you ask someone to do you a favor, like answer a question, it's not a good idea to make it really hard for them to help you.

I seem to read it fine once I open the attachment.
It is asking given we have ##f(x) = (2x-1)^4##. The curve meets the ##x##-axis at a point ##P## and the line on the graph is a tangent to the curve at the point ##Q(1,1)##.
Find the area of the region bounded by the curve, the ##x##-axis, and the tangent to the curve at ##Q##.

squenshl said:
I seem to read it fine once I open the attachment.
It is asking given we have ##f(x) = (2x-1)^4##. The curve meets the ##x##-axis at a point ##P## and the line on the graph is a tangent to the curve at the point ##Q(1,1)##.
Find the area of the region bounded by the curve, the ##x##-axis, and the tangent to the curve at ##Q##.

I cannot read it once I open the attachment.

Anyway, the PF standard is that the person wanting help takes the trouble to type out the problem and the work done so far, reserving attachments for things like diagrams and perhaps tables, etc. Read the Guidelines for more on this issue.

Ok I found the equation of the tangent curve in the standard way to get ##y=8x-7## which cuts the x-axis at ##x=\frac{7}{8}##.
To find P I found the x coordinate Of the turning point which is ##x=0.5## so now we have our bounds of our integral so we just calculate ##\int_{0.5}^{7/8} (2x-1)^4-8x+7 \; dx##.
I get ##0.5862##.

My problem is that I need to set the magnification for Chrome to such a high value that all other screens become useless.
And all that because you don't want to type out the problem statement, which is quite straightforward an clear.
Good thing you found the 0.5 by yourself, as well as the 7/8.
But in the figure I clearly see there is a part of the integral between 7/8 and 1
And I also see the integral between 0.5 and 7/8 is not how you describe it...

All this would have come out if you re-typed the problem statement and formulated what to do to find the answer in your own words, which is why using the template is so useful that it is compulsory in PF

squenshl said:
Member warned to type the problem statement, not just post an image with type that is too small to read

See attached.

## The Attempt at a Solution

Ok so the first thing you want to do is find the equation of the tangent line which is done in the usual way to get ##y=8x-7##. This cuts the ##x## axis at ##x=\frac{7}{8}## so we know the upper limit of the integral to calculate.
Is ##P## a turning point if not how do I find it?
P is the point where the curve meets the x axis. Can you find it?

ehild said:
P is the point where the curve meets the x axis. Can you find it?
Yes it’s ##x=0.5##.

Good. That is the lower bound of the integral. What is the upper bound ?

BvU said:
Good. That is the lower bound of the integral. What is the upper bound ?
I’m thinking it’s where the tangent function cuts the ##x##-axis which is when ##8x-7=0##, or ##x=\frac{7}{8}##.

Oops no it isn’t that will be ##x=1##.

So that means I solve ##\int_{0.5}^{1} (2x-1)^4 -8x + 7 \; dx## in which case I get ##0.6##?

Don't jump to conclusions -- they aren't interesting anyway. Do you see a difference between the regions 0.5 -- 7/8 and 7/8 -- 1 ?

squenshl said:
So that means I solve ##\int_{0.5}^{1} (2x-1)^4 -8x + 7 \; dx## in which case I get ##0.6##?
As a kind of check:
Cut a piece of paper of 1 in x and 0.6 in y and compare its area with the shaded area

BvU said:
As a kind of check:
Cut a piece of paper of 1 in x and 0.6 in y and compare its area with the shaded area
Hold up.
I can just calculate ##\int_{0.5}^{1} (2x-1)^4 \; dx=0.1##.
Then calculate ##\int_{7/8}^{1} 8x-7 \; dx=0.0625##.
Now take the difference to get ##\frac{3}{80}=0.0375##.

As long as you see that there are two parts, things go fine -- and that's the important part of what you've learned from this exercise.
Now that you have an answer and are happy, try the next one !

## 1. What is a tangent line?

A tangent line is a straight line that touches a curve at only one point, and has the same slope as the curve at that point. It represents the instantaneous rate of change or slope of the curve at that point.

## 2. How do you find the equation of a tangent line?

To find the equation of a tangent line, you need to first find the slope of the curve at the desired point. This can be done by taking the derivative of the curve at that point. Then, you can use the point-slope form of a line to plug in the coordinates of the point and the slope to find the equation of the tangent line.

## 3. What is the relationship between tangent lines and limits?

Tangent lines and limits are closely related because the slope of a tangent line at a specific point is the same as the limit of the slope of the curve as it approaches that point. In other words, the slope of the tangent line is the limit of the slope of the curve as the distance between two points on the curve gets smaller and smaller.

## 4. Can a curve have multiple tangent lines at a single point?

No, a curve can only have one tangent line at a specific point. This is because a tangent line must touch the curve at only one point and have the same slope as the curve at that point. If there were multiple tangent lines at a single point, they would have different slopes and therefore would not be tangent lines.

## 5. How are tangent lines and derivatives related?

Tangent lines and derivatives are related because the derivative of a curve at a specific point is equal to the slope of the tangent line at that point. In other words, taking the derivative of a curve at a point gives you the slope of the tangent line at that point.

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