Consider the function 𝑓(𝑥)=𝑥4−4⋅𝑥2. Find the relevant curve sketching information and then select the correct graph from the list below.𝑥-intercepts: {−2,2} Enter your answers inside curly brackets, e.g. {1,2}𝑦-intercept: 0Relative minima: [𝑥,𝑦]= Enter your answer as an ordered pair inside square brackets, e.g. [1,2]. If there is more than one relative minimum, enter your answers inside curly brackets, e.g. {[1,2],[3,4]}.Relative maxima: [𝑥,𝑦]= Enter your answer as an ordered pair inside square brackets, e.g. [1,2]. If there is more than one relative maximum, enter your answers inside curly brackets
Question
Consider the function 𝑓(𝑥)=𝑥4−4⋅𝑥2. Find the relevant curve sketching information and then select the correct graph from the list below.𝑥-intercepts: {−2,2} Enter your answers inside curly brackets, e.g. {1,2}𝑦-intercept: 0Relative minima: [𝑥,𝑦]= Enter your answer as an ordered pair inside square brackets, e.g. [1,2]. If there is more than one relative minimum, enter your answers inside curly brackets, e.g. {[1,2],[3,4]}.Relative maxima: [𝑥,𝑦]= Enter your answer as an ordered pair inside square brackets, e.g. [1,2]. If there is more than one relative maximum, enter your answers inside curly brackets
Solution
The function given is 𝑓(𝑥)=𝑥^4−4⋅𝑥^2.
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To find the x-intercepts, we set 𝑓(𝑥) = 0 and solve for x. 𝑥^4−4⋅𝑥^2 = 0 𝑥^2(𝑥^2 - 4) = 0 𝑥^2 = 0 or 𝑥^2 = 4 So, 𝑥 = 0, -2, 2. Therefore, the x-intercepts are {−2, 0, 2}.
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The y-intercept is the value of the function at x = 0. So, 𝑓(0) = 0^4 - 4*0^2 = 0. Therefore, the y-intercept is 0.
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To find the relative minima and maxima, we first find the derivative of the function, 𝑓'(𝑥) = 4𝑥^3 - 8𝑥. We set this equal to zero and solve for x to find critical points. 4𝑥^3 - 8𝑥 = 0 4𝑥(𝑥^2 - 2) = 0 So, 𝑥 = 0, √2, -√2 are the critical points.
We then use the second derivative test to classify these points. The second derivative of the function is 𝑓''(𝑥) = 12𝑥^2 - 8. We substitute the critical points into the second derivative.
𝑓''(0) = -8 (which is less than 0, so x = 0 is a relative maximum) 𝑓''(√2) = 8 (which is greater than 0, so x = √2 is a relative minimum) 𝑓''(-√2) = 8 (which is greater than 0, so x = -√2 is a relative minimum)
Therefore, the relative minima are {[−√2, -2], [√2, -2]} and the relative maximum is [0, 0].
Similar Questions
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