For a function to be continuous at an x-value, say -17, you need to make sure two things line up:
The limit from the left equals the limit from the right.
This limit equals the functions value.
The left hand limit involves the first piece, f(x) = 20x + 1:
The right hand limit invovles the second piece, f(x) = -10x^2:
Since the two one-sided limits don't match, the function is not continuous at x=-17.
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Answer:
4) 6x
5) 2x +3
Step-by-step explanation:
We can work both these problems at once by finding an applicable rule.
where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.
This can be referred to as the <em>power rule</em>.
Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:
lim[h→0](f(x+h)-f(x))/h = 2ax +b
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4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.
5. The gradient of x^2 +3x +1 is 2x +3.
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If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.