Calculate the degree of unsaturation: the answer is 1. The molecule can have a carbon-carbon double bond, a carbonyl, or a ring.
If you have a double bond and an oxygen in the molecule, always look for a carbonyl, because carbonyl bands are strong and easy to find. Sure enough, the band at 1713 indicates a carbonyl. Another prominent band is the characteristic wide O-H stretch at about 3150; this band looks more like a carboxylic acid O-H stretch than an alcohol O-H stretch (see example 3 for IRs of alcohols). C-O stretches show up from 1320-1000.
In the NMR, since we suspect a carboxylic acid, look for a broad singlet in the region 10-13.2 ppm.
Indeed, the broad peak at 11.8 indicates a carboxylic acid proton: -CO2H.
The triplet at 1 ppm corresponds to a methyl group next to a methylene (-CH2-) group; the sextet at 1.7 ppm is a methylene group next to a total of 5 hydrogens; the triplet at 2.3 ppm is a methylene group next to a methylene group. Therefore, it looks like the molecule is straight-chain:
The structure correlates with the NMR spectrum like this: