+1904 Estimate the instantaneous rate of change for the function f(x)=e^x at x=0, then use this to find an equation for the tangent line. Check your answer by graphing the function and it's tangent line in Desmos. Print out and attach a copy of your graph.
I have a worksheet in my calculus class due on Monday, April 9, 2018 and need help with this question. Anyone who knows the answer and can show step-by-step instructions, I would really appreciate it. Thanks.
+118687 Estimate the instantaneous rate of change for the function f(x)=e^x at x=0, then use this to find an equation for the tangent line. Check your answer by graphing the function and it's tangent line in Desmos. Print out and attach a copy of your graph.
f(x)=e^x
f'(x)=e^x
f'(0)=e^0=1
So the instantaneous rate of change at x=0 is 1, this is also the gradient of the tangent to the curve at x=0 as that is what the instantaneous rate of change IS
When x=0
f(0) = e^0 = 1
so what is the equation of the line through (0,1) with a gradient of 1
y=mx+b
y=1x+1
y=x+1
You can do the desmos graph I expect :)
+1904 Thanks for the help. I way over thought this problem. y=mx+b makes total sense. I feel really dumb. Now I can finish my homework. I do know how to use Desmos.
