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January 29, 2022, 06:34:48 am

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#### Lachlaaaaaaan

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« on: March 21, 2021, 09:48:32 pm »
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This question Is from the Jacaranda 12 Methods Textbook, I'm really struggling to understand part d and e of the question and how to solve it. Any help would be appreciated

Consider the graph defined by the function F(x)= (x^2)(e^(1-x))
a. differentiate to find f(x)'
b. Determine the coordinates of stationary points
c. Sketch the function
d. The point P(k, f(k)) lies on the curve. Determine the gradient of the line joining Point P to the Origins O.
e. Hence determine the coordinates of any non-stationary point(s) on the curve where the tangent passes through the origin, and determine the equation of the tangent.

#### fun_jirachi

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« Reply #1 on: March 22, 2021, 01:01:48 am »
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Welcome to the forums!

For part d), just use $m = \frac{y_2-y_1}{x_2-x_1}$ ie. the gradient formula. Since you're calculating the gradient of line joining P and the origin the gradient should just be $\frac{f(k)}{k}$.

For part e), a tangent line to a curve $f(x)$ at $x = k$ has gradient $f'(k)$. However, from part d) we also have that this line must also have gradient $\frac{f(k)}{k}$. Therefore, you should be looking for values of x where $f'(x) - \frac{f(k)}{k} = 0$ Are there any points that are not common with the points you found in part b) that satisfy this? Can you then put the pieces together to find the equation of such a tangent (use point-gradient form of the line).

Hope this helps
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