This week we learned about finding derivatives using the 4 step process:
1. Find f(x+h)
2. Write f(x+h)-f(x) and simplify
3. Write f(x+h)-f(x)
h
4. Take lim f(x+h)-f(x)
h→0 h
A problem that I struggled with at first on the sect. 3.1 (part 1) worksheet was number 7. The problem read f(x)=x^3+x; (0,f(0)). The first thing I did was replace all the x's with (x+h) (step 1) and I got (x+h)^3+(x+h). I then foiled out the (x+h)^3 and got f(x)=x^3+2x^2h+xh^2+xh^2+2xh^2+h^3+x+h. I combined like terms and got f(x)=x^3+2x^2h+4xh^2+h^3+x+h. Next, I subtracted f(x) from it (step 2) and got f(x)=x^3+2x^2h+4xh^2+h^3+x+h-(x^3+x). After canceling out terms, I ended up with f(x)=2x^2h+4xh^2+h^2+1. Then I divided it all by h (step 3) so it canceled an h from every term in the numerator and got f(x)=2x^2+4xh+h^2+1. Lastly, I look the limit of 2x^2+4xh+h^2+1 as h approaches 0 (step 4). I plugged in 0 for all h's and ended up with f'(x)=2x^2+1. Then I plugged 0 into f(x) so it read (0)^3+0=0. I knew my coordinate to find the slope of the tangent line was (0,0). To find the tangent line I plugged 0 into the f'(x) function so it read 2(0)^2+1=1. After finding out that the slope was 1, I plugged that and the coordinate (0,0) into slope intercept form (y=mx+b) in order to find the y-intercept. I got 0=1(0)+b so b=0. From there, I knew my equation for the tangent line was y=x.
We also did an activity where we found the slope of multiple tangent lines and eventually plotted the points to make the graph of the derivative. I noticed that the zeros of the new graph were at the same spots as when the slope was 0 on the original graph. I also noticed that the original graph was that of a cubic function, and its derivative was that of a quadratic.
Another activity we did was Lab 6: relationship between a function and its derivative. The most important thing I learned from this lab was the conjecture that the intervals of when f(x) is rising is the same as when f'(x) is above the x-axis, and the intervals of when f(x) is falling is the same as when f'(x) is below the x-axis. This was proven multiple times through desmos and different questions in the packet.
1. Find f(x+h)
2. Write f(x+h)-f(x) and simplify
3. Write f(x+h)-f(x)
h
4. Take lim f(x+h)-f(x)
h→0 h
A problem that I struggled with at first on the sect. 3.1 (part 1) worksheet was number 7. The problem read f(x)=x^3+x; (0,f(0)). The first thing I did was replace all the x's with (x+h) (step 1) and I got (x+h)^3+(x+h). I then foiled out the (x+h)^3 and got f(x)=x^3+2x^2h+xh^2+xh^2+2xh^2+h^3+x+h. I combined like terms and got f(x)=x^3+2x^2h+4xh^2+h^3+x+h. Next, I subtracted f(x) from it (step 2) and got f(x)=x^3+2x^2h+4xh^2+h^3+x+h-(x^3+x). After canceling out terms, I ended up with f(x)=2x^2h+4xh^2+h^2+1. Then I divided it all by h (step 3) so it canceled an h from every term in the numerator and got f(x)=2x^2+4xh+h^2+1. Lastly, I look the limit of 2x^2+4xh+h^2+1 as h approaches 0 (step 4). I plugged in 0 for all h's and ended up with f'(x)=2x^2+1. Then I plugged 0 into f(x) so it read (0)^3+0=0. I knew my coordinate to find the slope of the tangent line was (0,0). To find the tangent line I plugged 0 into the f'(x) function so it read 2(0)^2+1=1. After finding out that the slope was 1, I plugged that and the coordinate (0,0) into slope intercept form (y=mx+b) in order to find the y-intercept. I got 0=1(0)+b so b=0. From there, I knew my equation for the tangent line was y=x.
We also did an activity where we found the slope of multiple tangent lines and eventually plotted the points to make the graph of the derivative. I noticed that the zeros of the new graph were at the same spots as when the slope was 0 on the original graph. I also noticed that the original graph was that of a cubic function, and its derivative was that of a quadratic.
Another activity we did was Lab 6: relationship between a function and its derivative. The most important thing I learned from this lab was the conjecture that the intervals of when f(x) is rising is the same as when f'(x) is above the x-axis, and the intervals of when f(x) is falling is the same as when f'(x) is below the x-axis. This was proven multiple times through desmos and different questions in the packet.