Anonymous From: - Posts: - Votes: - |
Please advise & share with me how to solve this question. Thanks a lot. Find a value for the constant k, if possible, that will make the function continuous everywhere 7x – 2 x≤ 1 f(x) = kx2 x>1 kx2 refers to k square(x). Have a nice day. Thank you. |

EliteBrainy From: US Posts: 12 Votes: 36 |
I believe k is 5... because for the function to be continuous, 7x-2 must be equal to kx^2 at x=1 |

Anonymous From: - Posts: - Votes: - |
Since you want the function to be continous, you have to find the point where the two graphs intersect. The point of intersection is given by the constraint. the graphs intersect at x=1 so 7x-2 = kx^2, when x=1 sub in the one for x and solve for k 7(1) - 2 = k(1)^2 7-2 = k 5 = k so f(x)={7x-2; x1}buy viagra nigeria list of reputable canadian pharmacies canada viagra generic list of reputable canadian pharmacies generic viagra for sale cheap viagra without prescription buy red viagra viagra without a doctor prescription usa does the generic viagra work |

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