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Consider the function ` f(x) = 4 -3 x^2 ` on the interval ` [ -2 , 3 ] `.
Find the average or mean slope of the function on this interval, i.e.
` \frac ( f(3) - f(-2) )( 3 - (-2) ) = `
By the Mean Value Theorem, we know there exists a ` c ` in the open interval ` (-2, 3) ` such that ` f'( c) ` is equal to this mean slope. For this problem, there is only one ` c ` that works. Find it.
` \frac ( f(3) - f(-2) )( 3 - (-2) ) = `
By the Mean Value Theorem, we know there exists a ` c ` in the open interval ` (-2, 3) ` such that ` f'( c) ` is equal to this mean slope. For this problem, there is only one ` c ` that works. Find it.
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Box 1: Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter DNE for Does Not Exist, oo for Infinity
-3
Box 2: Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter DNE for Does Not Exist, oo for Infinity
0.5