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I have a Calculus derivative problem.?
So...the function is f(x)=e^x - 2x^2
with the given point (0) i have found out that the tanget line is y=x+1
When drawing the two graphs(the function and the tangent line) I have to draw 2 exact points on each graph. I got two down by placing the tangent point (1,0) [so now i have 1 on the function and one on the line] and the (-1,0) point on the tangent line. The only problem is I can't find a second exact point on the graph of the function itself (without any decimals, functions such as *square root* or *natural log* are allowed in the "name" of the point...as long as one of them is a whole number).
Thank you for your help.
I know what the derivative is.....and I don't need the derivative.....and I know the slope of the tangent line.....I was talking about the tangent line equation which i obviously got right.
The only thing that I need is to LABEL a point on the GRAPH of the FUNCTION that is f(x)=e^x -2x^2
1 Answer
- 1 decade agoFavorite Answer
You are over thinking this question, aside from that, your derivative is wrong.
f(x) = e^x -2x^2
f'(x) = e^x -4x
f'(0) = e^0 -4*0 = e^0 = 1
The slope of the tangent line is 1, and the y intercept is 0. The line is therefore f(x) = x
To plot it, just pick any x and the f(x) is the exact same number.
Here are he numbers you need
x f(x) tangent line
1.0e+004 (multiply each by 10000)
-0.0010 -0.0010 -0.0200
-0.0008 -0.0008 -0.0121
-0.0006 -0.0006 -0.0062
-0.0003 -0.0003 -0.0022
-0.0001 -0.0001 -0.0002
0.0001 0.0001 0.0001
0.0003 0.0003 0.0006
0.0006 0.0006 0.0197
0.0008 0.0008 0.2266
0.0010 0.0010 2.1826