Calculus please help- The tangent line to the graph of g(x) at x = −1 is given by y = 4(x + 1) + 7.?
Please help me understand this problem thanks, 11 points.
The tangent line to the graph of g(x) at x = −1 is given by y = 4(x + 1) + 7. Suppose
F(x) = x^3 g(x)
Find F ' (−1).
Update:Possibilities:
(a) 25
(b) 17
(c) −4
(d) 4
(e) −17
6.
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Verified answer
We know that the tangent line to a point on some other curve must pass through that point and must have the same slope as the value of the derivative at that point.
With that being said, the corresponding y-value to x = -1 is y = 4(-1 + 1) + 7 = 7. Thus, g(x) passes through (-1, 7) and g(-1) = 7. Then, since the slope of the tangent line is 4 (you can see this by expanding and comparing to slope-intercept form), g'(1) = 4.
Finally, using the Product Rule:
F'(x) = 3x^2*g(x) + x^3*g'(x).
So, F'(-1) = 3(-1)^2*g(1) + (-1)^3*g'(1) = 3g(1) - g'(1) = 3(7) - 4 = 17.
I hope this helps!