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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.

Update:

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.

Update 2:

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

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  • 1 decade ago
    Favorite 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

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