Calculus Help

Question

Let ƒ(x)=e^−(x+1)^2.
Draw a graph of the function on the interval [-3,3] and estimate from the graph whether ƒ′(−1) is positive or negative or zero, and whether ƒ′′(−1) is positive, negative or zero.

Answer 1

f(x) = e^{-(x+1)^2}

f'(x) = -2(x+1)e^{-(x+1)^2}

f''(x) = 4(x+1)²e^{-(x+1)^2} - 2e^{-(x+1)^2}

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f'(-1) = -2(-1+1)e^{-(-1+1)^2} = 0

f''(-1) = 4(-1+1)2e^{-(-1+1)^2} - 2e^{-(-1+1)^2} = 0 - 2e⁰ = -2

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as for the graph: http://www.wolframalpha.com/input/?i=plot+y+%3D+e%5E(-(x%2B1)%5E2),+-3%3C%3Dx%3C%3D3

Answer 2

Hi Anthony,

 

Visit the following graph to see what is happening. Use the "slider" for "c" moving it to -1. You will see that at x=-1, the tangent line has a slope of 0. As the slider moves from -3 to 3, but in the vicinity of (-1,1) the tangent line is decreasing in slope, i.e. the 2nd derivative is negative. The curve lies under the tangent line so f is concave downward. You would have to actually find the 2nd derivative to locate the points of inflection.  

 

https://www.desmos.com/calculator/zls3tdpptt